Rant Thread
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Rant Thread
Just rant about what you wanna rant about.
I'll begin.
I can't wait till I move, this appartment is a hole.
I'll begin.
I can't wait till I move, this appartment is a hole.
White Linen Posts : 105
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Re: Rant Thread
My stupid brother keeps slowing down my internets!! It feels like I'm back to having dial up.
Miss_Ganja Posts : 146
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I just moved upstairs in my townhouse, but we got robbed recently so I am less happy. Also i'm broke and ikea is expensive bitches.
evolsine Posts : 110
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my mathematics course is getting extremely confusing. Anyone pro with Logarithmic series and recurrance relations that you have to factorize into a quadratic formula to get part of a solution to get the equation needed to solve the original non homogeneous equation?! ANYONE!?
Menyth Posts : 56
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I'll take a shot at it. Can it wait for about half a day, though?
Layra Posts : 65
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Layra wrote:I'll take a shot at it. Can it wait for about half a day, though?
I've managed to solve one problem. But there's this one where you have to use a check to see if it's correct. Something strange. But it can wait, it's due Thursday. (24 hours from now)
Here's the equation.
Says "Give a complete solution, including the check, for the first order IVP"....
Also, I've found the answer to this equation, do you think you can explain it though? Coz I just followed the steps on how to do it and I think it's correct, but I just don't get the logical explanation to it. If you could, it wold greatly help my understanding
Menyth Posts : 56
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What is 3_n supposed to mean? Is that just 3 times n, or is that actually supposed to be a subscript?
Layra Posts : 65
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If it is supposed to be 3 times n, then you must have done something wrong, because I don't think your solution works. Even evaluating at 0 gives you problems.
Here's what I did: first, you solve the homogeneous equation H_n  4H_{n1} = 0. The obvious solution: H_n = a4^n for constant a. We'll find a later.
Now we go back to the inhomogeneous equation T_n  4T_{n1} = 3n+2. We need to find a single inhomogeneous solution, which I'll call I_n. The 3n indicates that we can try something stupid like I_n = n. That would give us I_n  4I_{n1} = n  4((n1)) on the left side, which expands to 3n  4. Not quite what we want. However, note that replacing I_n = n with I_n = n + b gives that I_n  4I_{n1} = 3n  4  3b. So we set b = 2 and get that for I_n = n  2, I_n  4I_{n1} = 3n+2.
Since H_n is a solution to the homogeneous equation, I_n+H_n is a solution to the inhomogeneous equation.
From here we just need to match the initial condition: T_0 = 4. I_0 = 2, and H_0 = a, so we set a = 2 and get T_n = 2(4^n)  n  2.
I have no idea what you did. But you still have evaluation at 0 problems.
Here's what I did: first, you solve the homogeneous equation H_n  4H_{n1} = 0. The obvious solution: H_n = a4^n for constant a. We'll find a later.
Now we go back to the inhomogeneous equation T_n  4T_{n1} = 3n+2. We need to find a single inhomogeneous solution, which I'll call I_n. The 3n indicates that we can try something stupid like I_n = n. That would give us I_n  4I_{n1} = n  4((n1)) on the left side, which expands to 3n  4. Not quite what we want. However, note that replacing I_n = n with I_n = n + b gives that I_n  4I_{n1} = 3n  4  3b. So we set b = 2 and get that for I_n = n  2, I_n  4I_{n1} = 3n+2.
Since H_n is a solution to the homogeneous equation, I_n+H_n is a solution to the inhomogeneous equation.
From here we just need to match the initial condition: T_0 = 4. I_0 = 2, and H_0 = a, so we set a = 2 and get T_n = 2(4^n)  n  2.
I have no idea what you did. But you still have evaluation at 0 problems.
Layra Posts : 65
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My mind died. I dun get.
I go to a liberal arts college. Math goes in one ear and out the other.
I go to a liberal arts college. Math goes in one ear and out the other.
DanVeytia Posts : 73
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Oh, wait, is that second white bit a different problem from the first? 'Cause that would make more sense, but then I have no idea what you're supposed to be doing in the second bit. Where does the 14 come from?
Layra Posts : 65
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Yes, Sorry, the later equation has absolutely no relation to the first equation. It was just there to see if you could explain how I got to 14n^2 log(n). Apparently you have to take the modolus of each segment of the original equation (8n^2 long(n) + n^2 + 3n log(n) +2n). Here's my working out for that equation:
(8n^2 log(n) + n^2 + 3n log(n) +2n) <= 8n^2 log(n) + n^2 + 3nlog(n) + 2n
=8n^2 log(n) + n^2 log(n) +3n^2 log(n) + 2n^2 log(n)
=14n^2 log(n)
The theorm used is called the "Triangle inequality : a + b <= a + b" if that helps you understand. Because I don't see the logic behind it, was wondering if you knew :S
Im still confused about the original though, I totally don't get all that substituting different values from nowhere D:
Oh and the T_n is a mistake, it's supposed to be T*n. Yea. Sorry,
(8n^2 log(n) + n^2 + 3n log(n) +2n) <= 8n^2 log(n) + n^2 + 3nlog(n) + 2n
=8n^2 log(n) + n^2 log(n) +3n^2 log(n) + 2n^2 log(n)
=14n^2 log(n)
The theorm used is called the "Triangle inequality : a + b <= a + b" if that helps you understand. Because I don't see the logic behind it, was wondering if you knew :S
Im still confused about the original though, I totally don't get all that substituting different values from nowhere D:
Oh and the T_n is a mistake, it's supposed to be T*n. Yea. Sorry,
Menyth Posts : 56
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For the second part, you're going to have to explain again what the problem itself says, because your bit about the modulus is nonsense as far as I can see. Why is the modulus of 3nlog(n) equal to 3n^2log(n)? Normally, modulus just means "absolute value", i.e. you get rid of any negative signs.
In fact, I'm almost sure that whatever you did is wrong because if you evaluate at 1, 14*1^2log(1) = 0, while 8*1^2log(1) + 1^2+3*1*log(1) + 2*1 = 3, and 3 > 0. So yeah, your definition of modulus is junk.
And for the first part, you're looking at T*n? So it's supposed to read T*n4T*(n1) = 3*n+2? And by * do we mean times, or exponential? Because it it's times, then you're just looking at T(3n+1) = 3n+2, which gives that T = (3n+2)/(3n+1). Which you should know how to do yourself. The problem only becomes interesting at all if it's not times.
Yeah, you're probably going to have to give me the original statements of the problems, because I have no idea what you're trying to do here.
In fact, I'm almost sure that whatever you did is wrong because if you evaluate at 1, 14*1^2log(1) = 0, while 8*1^2log(1) + 1^2+3*1*log(1) + 2*1 = 3, and 3 > 0. So yeah, your definition of modulus is junk.
And for the first part, you're looking at T*n? So it's supposed to read T*n4T*(n1) = 3*n+2? And by * do we mean times, or exponential? Because it it's times, then you're just looking at T(3n+1) = 3n+2, which gives that T = (3n+2)/(3n+1). Which you should know how to do yourself. The problem only becomes interesting at all if it's not times.
Yeah, you're probably going to have to give me the original statements of the problems, because I have no idea what you're trying to do here.
Layra Posts : 65
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Don't worry about the second equation, I checked with my lecturer, if all the terms are positive, I don't have to take the modulus of each term.
I know the sequence to this now. It's called the "lazy c" approach, or something like that.
Once all the terms are positive, you must identify the dominant term.
With the dominant term, you 'dominate' the other terms; as in assimilate it into the terms.
so say the terms are n^2 + 5n +n log(n) + n^2 log (n) for all n>=2. The dominant term would be n^2 log (n) because it'll give the largest value.
Once 'dominated' the equation should look like this: n^2 log (n) + 5n^2 log (n) + n^2 log(n) + n^2 log(n). Then its just a matter for adding it all up (which will be 8n^2 log(n)). From that I write c=8, M=2 (because of n>=2)
He corrected that question and I got it right, I just don't understand why I have to do that.
The question is "Show that 8n^2 log(n) + n^2 + 3n log(n) + 2n is an element of 'Big O'(n^2 log(n))"
yea. Not much help, you don't have to explain it, I just hoping someone would so I could know it with confidence.
As for the other equation, I'm sorry, it was another typo . The equation is "T_n  4T_(n1) = 3n + 2", T_0 = 4 (where n >= 1). Once again i'll mention it's asking "Give a complete solution, including the check, for the first order IVP".
I figured something out about it, on the RHS of that equation, it's a linear equation (mx+c) and in my text book it says 'Use An + B as a general solution for all linear equations on the RHS for nonhomogenous equations' the same goes for all RHS such as 'n' or '3n2' etc.
From there, I subsituted the General Solution (An + B) into T_n  4T_(n1) so I get An + B  4(A(n1) + B).
Expanded it leads to:
An + B 4A(n1)  4B
therefore: 3An + (4A  3B)
With that I make 3An = 3 (as 3 is the coefficent of n in the RHS of the equation) so it equals 3A = 3 which leads to A=1 and B .... I'm not sure what I should make (4A  3B) equal to so I can find the solution to that. Yea D:
Don't worry if this is too much effort or any at all, don't want to bother you D: It might be different to you as well, coz in this mathematics we commonly use log base 2 or log base 'e'. So it might be a bit more obscure to you. But you are right log (0) is not valid and log (1) = 0, that is why we use n>=2 (you know, just to clarify.)
I know the sequence to this now. It's called the "lazy c" approach, or something like that.
Once all the terms are positive, you must identify the dominant term.
With the dominant term, you 'dominate' the other terms; as in assimilate it into the terms.
so say the terms are n^2 + 5n +n log(n) + n^2 log (n) for all n>=2. The dominant term would be n^2 log (n) because it'll give the largest value.
Once 'dominated' the equation should look like this: n^2 log (n) + 5n^2 log (n) + n^2 log(n) + n^2 log(n). Then its just a matter for adding it all up (which will be 8n^2 log(n)). From that I write c=8, M=2 (because of n>=2)
He corrected that question and I got it right, I just don't understand why I have to do that.
The question is "Show that 8n^2 log(n) + n^2 + 3n log(n) + 2n is an element of 'Big O'(n^2 log(n))"
yea. Not much help, you don't have to explain it, I just hoping someone would so I could know it with confidence.
As for the other equation, I'm sorry, it was another typo . The equation is "T_n  4T_(n1) = 3n + 2", T_0 = 4 (where n >= 1). Once again i'll mention it's asking "Give a complete solution, including the check, for the first order IVP".
I figured something out about it, on the RHS of that equation, it's a linear equation (mx+c) and in my text book it says 'Use An + B as a general solution for all linear equations on the RHS for nonhomogenous equations' the same goes for all RHS such as 'n' or '3n2' etc.
From there, I subsituted the General Solution (An + B) into T_n  4T_(n1) so I get An + B  4(A(n1) + B).
Expanded it leads to:
An + B 4A(n1)  4B
therefore: 3An + (4A  3B)
With that I make 3An = 3 (as 3 is the coefficent of n in the RHS of the equation) so it equals 3A = 3 which leads to A=1 and B .... I'm not sure what I should make (4A  3B) equal to so I can find the solution to that. Yea D:
Don't worry if this is too much effort or any at all, don't want to bother you D: It might be different to you as well, coz in this mathematics we commonly use log base 2 or log base 'e'. So it might be a bit more obscure to you. But you are right log (0) is not valid and log (1) = 0, that is why we use n>=2 (you know, just to clarify.)
Menyth Posts : 56
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You look like you're on the right track with the first part. You want (4A3B) = 2, since that's the constant term. Since you already know A, you substitute and solve for B.
This gives you a solution for T_n4T_(n1) = 3n+2. Let's call it F_n.
The next step would be to solve the more general equation T_n4T_(n1) = 0 to get a solution. We separate T_n and T_(n1) to get T_n = 4T_(n1), with the obvious solution being 4^n; call this G_n.
Note that F_n+kG_n for any k is a solution to T_n4T_(n1) = 3n+2, since G_n doesn't add anything to the right side. So now you have the general solution k4^nn2. From here you need to solve for the initial value, i.e. plug in 0 and see what k needs to be.
That's the general method for solving these types of equations (specifically linear difference equations). First, find a solution to the full equation, via various tricks or divination or finding a compsci major and beating him until he capitulates (I recommend the last method). Then, find a solution to the homogeneous equation, i.e. set the right side to 0 and solve. This will give you a function such that any constant multiple of that function is a solution.
Then you take your first solution and add an unknown constant times the second solution, and then evaluate at your initial value to find what the unknown constant has to be.
This gives you a solution for T_n4T_(n1) = 3n+2. Let's call it F_n.
The next step would be to solve the more general equation T_n4T_(n1) = 0 to get a solution. We separate T_n and T_(n1) to get T_n = 4T_(n1), with the obvious solution being 4^n; call this G_n.
Note that F_n+kG_n for any k is a solution to T_n4T_(n1) = 3n+2, since G_n doesn't add anything to the right side. So now you have the general solution k4^nn2. From here you need to solve for the initial value, i.e. plug in 0 and see what k needs to be.
That's the general method for solving these types of equations (specifically linear difference equations). First, find a solution to the full equation, via various tricks or divination or finding a compsci major and beating him until he capitulates (I recommend the last method). Then, find a solution to the homogeneous equation, i.e. set the right side to 0 and solve. This will give you a function such that any constant multiple of that function is a solution.
Then you take your first solution and add an unknown constant times the second solution, and then evaluate at your initial value to find what the unknown constant has to be.
Layra Posts : 65
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The second problem isn't actually that difficult to do. First, having taken the modulus of everything (i.e. made everything positive), you look at the leading term. If you've ordered the terms correctly, the leading term will grow faster than all the other terms (ignoring coefficients). In this case, n^2log(n) grows faster than n^2 because log(n) grows, and n^2log(n) grows faster than nlog(n) because n grows, and n^2log(n) grows faster than n because nlog(n) grows. So n^2log(n) is your leading term.
To show that 8n^2log(n)+blah is O(n^2log(n)), you ditch the coefficient and see that yes, n^2log(n) = n^2log(n), so 8n^2log(n) is O(n^2log(n)), and hence 8n^2log(n)+blah is O(n^2log(n)).
If you want to be more precise, here's how you're actually supposed to calculate it, in case you didn't order the terms correctly:
You take (8n^2log(n)+blah)/n^2log(n), and take the limit as n goes to infinity. (8n^2log(n)+blah)/n^2log(n) = 8n^2log(n)/n^2log(n) + blah/n^2log(n). blah/n^2log(n) goes to 0 as n goes to infinity, and 8n^2log(n)/n^2log(n) goes to 8. Since this is finite and nonzero, we get that 8n^2log(n)+blah is in O(n^2log(n)).
To show that 8n^2log(n)+blah is O(n^2log(n)), you ditch the coefficient and see that yes, n^2log(n) = n^2log(n), so 8n^2log(n) is O(n^2log(n)), and hence 8n^2log(n)+blah is O(n^2log(n)).
If you want to be more precise, here's how you're actually supposed to calculate it, in case you didn't order the terms correctly:
You take (8n^2log(n)+blah)/n^2log(n), and take the limit as n goes to infinity. (8n^2log(n)+blah)/n^2log(n) = 8n^2log(n)/n^2log(n) + blah/n^2log(n). blah/n^2log(n) goes to 0 as n goes to infinity, and 8n^2log(n)/n^2log(n) goes to 8. Since this is finite and nonzero, we get that 8n^2log(n)+blah is in O(n^2log(n)).
Layra Posts : 65
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What the fuck does log mean.
DanVeytia Posts : 73
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I actually understand now.
I managed to get the complete solution of T(n) =6(4)^n  n  2.
I just got the homogeneous solution and inhomogeneous solution and added them together (S_p (n) = H_n + S_n).
Submitted my assignment. Lets see how I went next week I'll post the results
I managed to get the complete solution of T(n) =6(4)^n  n  2.
I just got the homogeneous solution and inhomogeneous solution and added them together (S_p (n) = H_n + S_n).
Submitted my assignment. Lets see how I went next week I'll post the results
Menyth Posts : 56
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Logarithm.......DanVeytia wrote:What the fuck does log mean.
If my memory serves me correctly.
logÂ·aÂ·rithm
â€‚ â€‚/ËˆlÉ”gÉ™ËŒrÉªÃ°É™m, ËŒrÉªÎ¸, ËˆlÉ’gÉ™/ Show Spelled[lawguhrithuhm, rith, loguh] Show IPA
â€“noun Mathematics .
the exponent of the power to which a base number must be raised to equal a given number; log: 2 is the logarithm of 100 to the base 10 (2 = log 10 100).
Use logarithms in a Sentence
See images of logarithms
Search logarithms on the Web
Origin:
1605â€“15; < NL logarithmus < Gk lÃ³g ( os ) log + arithmÃ³s number; see arithmetic
White Linen Posts : 105
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Shto etyo?
evolsine Posts : 110
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Thanks WL. But I'm never taking math again though. I can skip it when I'm in college.
DanVeytia Posts : 73
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Don't know if you guys been following. But the gaia f73 has been infected with some chrisEPIC schmuck. I say we need to smoke him out. Or burn. Or whatever floats your perverted yacht on the sea of disgust.
I really want to crush him, arrogant emo stooge thinks he can come in and start 'correcting' the definition of music.
I really want to crush him, arrogant emo stooge thinks he can come in and start 'correcting' the definition of music.
Menyth Posts : 56
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Ya srsly. Any ideas?Menyth wrote:Don't know if you guys been following. But the gaia f73 has been infected with some chrisEPIC schmuck. I say we need to smoke him out. Or burn. Or whatever floats your perverted yacht on the sea of disgust.
I really want to crush him, arrogant emo stooge thinks he can come in and start 'correcting' the definition of music.
DanVeytia Posts : 73
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Those punks play games, just strip the game and he'll run.
White Linen Posts : 105
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he posted the link to his soundcloud, everything he makes is fucking awful lol
eskae Posts : 124
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My ears are bleeding from that Ducky song. Fuck my life.
DanVeytia Posts : 73
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yeah, that was one of the worst things ive heard in a really long time
eskae Posts : 124
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Oh man. Cypher's gonna blow a shit.
DanVeytia Posts : 73
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he lives for this shit man its gonna be great im hella excited LOL! may god have mercy on that poor child's soul
eskae Posts : 124
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eskae wrote:he lives for this shit man its gonna be great im hella excited LOL! may god have mercy on that poor child's soul
To be brutally honest, I actually do hope something bad happens to the kid. I can't stand arrogance. I don't know how I managed to stay with f.73 when it was filled with music elitests. But then I realized, I myself am one.. so... I guess I can stand enough arrogance from others, enough to not label me a hypocrite
Menyth Posts : 56
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Being an elitist is fine; being a complete jackass on key points is not.
evolsine Posts : 110
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I actually think I railed on you when you first popped on f.73..... Not sure thought.Menyth wrote:eskae wrote:he lives for this shit man its gonna be great im hella excited LOL! may god have mercy on that poor child's soul
To be brutally honest, I actually do hope something bad happens to the kid. I can't stand arrogance. I don't know how I managed to stay with f.73 when it was filled with music elitests. But then I realized, I myself am one.. so... I guess I can stand enough arrogance from others, enough to not label me a hypocrite
White Linen Posts : 105
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LOL! Dino moved the thread to the CB.
DanVeytia Posts : 73
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fucking tired of constantly being depressed
eskae Posts : 124
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Life has no meaning. Work, sleep. video game, produce, whatever
evolsine Posts : 110
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got da murda gloves an a fo' five finna do werk bwaiiii
eskae Posts : 124
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That used to be my case.eskae wrote:fucking tired of constantly being depressed
White Linen Posts : 105
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I just realize that with the meds i'm taking I can't do any fun drugs. DAMNIT!
White Linen Posts : 105
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MAOI inhibitors n shit? but look at it this way... its better to be sober and happy, than pissed off at the world on acid lol
eskae Posts : 124
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An SSRI and an anti psychotic.
E= Would do nothing and risk serotonin syndrome
LSD= The antipsychotic and SSRI would cancel pretty much all of it.
Shrooms=same thing.
Speed= You only get the body buzz, no euphoria.
They don't prescribe MAOI's that much anymore because of all the contraints and the diet you have to get on. MAOI's are pretty old...
Yeah I know that! I haven't been THIS happy and motivated in ages. I was just saying.
I just like to disconnect 23 times a year, it's always fun.
E= Would do nothing and risk serotonin syndrome
LSD= The antipsychotic and SSRI would cancel pretty much all of it.
Shrooms=same thing.
Speed= You only get the body buzz, no euphoria.
They don't prescribe MAOI's that much anymore because of all the contraints and the diet you have to get on. MAOI's are pretty old...
Yeah I know that! I haven't been THIS happy and motivated in ages. I was just saying.
I just like to disconnect 23 times a year, it's always fun.
White Linen Posts : 105
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i think my friend has a similar condition and medications. he took ayuhuasca and nothing happened, says hes tried acid a lot of times and never tripped.. cant imagine that. but yeah i should definitely look into some medications or something i feel like im finna snap any day now
eskae Posts : 124
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I take citalopram and quetiapine regularly. For the first 2 months try to have a mild antipsychotic(the new ones, the old shit they used to give is AWFUL for you) and maybe some anxiety meds.
I find it's best suited for me as I sometimes have extreme anxiety as well.
And if you think that it would be best for you do it man, best decision I took in the last 6 years. ANd don't forget that you need to find the right meds for you so don't hesitate to tell your psychiatrist if a med doesn't feel good.
I find it's best suited for me as I sometimes have extreme anxiety as well.
And if you think that it would be best for you do it man, best decision I took in the last 6 years. ANd don't forget that you need to find the right meds for you so don't hesitate to tell your psychiatrist if a med doesn't feel good.
White Linen Posts : 105
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thanks for the advice man ya ive felt so fucking hopeless thru all this. i feel like ive been subconsciously suicidal which is why i do werk in the ghettos now and hang with dangerous people. spend so much time on gaia and online in general because i hate meeting new people now yet always feel alone lol. realized it earlier today, felt weird to realize that im pathetic. if anything meds are worth a shot, if i still hate the world ill just leave my legacy of footprints in the form of beats and possibly a book, my life been so fucked it'd make a great movie lol
eskae Posts : 124
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Location : Vallejo, CA
Re: Rant Thread
Do you know there is a correlation between internet usage/computer usage and depression? Some study shown that the 90's, midlate 80's kids that spend alot of their time online and more likely to be depressed and antisocial?
IDK, i used to have mild depression when I was like... 10? Didn't do any meds but what I found helped was friends, hookers, alchohol. Just friends
Find something to do, a purpose; a job or something you enjoy doing. Try to limit your time online, see how that goes for a week or something
IDK, i used to have mild depression when I was like... 10? Didn't do any meds but what I found helped was friends,
Find something to do, a purpose; a job or something you enjoy doing. Try to limit your time online, see how that goes for a week or something
Menyth Posts : 56
Join date : 20100503
Age : 27
Location : Melbourne
Re: Rant Thread
i try to make music and suck at it. lost all my old friends over some shitty drama and now i dont kick it with trustworthy people anymore.. and yea i see how it is, you want me to limit my time online but ya trust ive tried all those things. i think id love to have a job, but at the same time i had a job before doing 40+ hours a week and i ended up breaking down and had such a bad anxiety attack that my heart felt like it was barely beating and was gonna stop, and i couldnt breathe... and things have gotten worse since then so no idea how that would go lol. but shit its almost gotten to the point where im used to being fucked in the head. the only problem is that more bad things keep happening and i still gotta deal with the old ones. murder seemed like the appropriate solution for a long time, was a point where every night id have such vivid dreams of killing people. always the same few people..
ive heard a lot of older people say that the early 20s are the hardest part in anyones life, so i been trying to keep to the belief that itll get better on its own. who knows tho.. lol. i try to work out as much as possible but its all for hate and violence. just dont kno anymore.. hope the pills help
ive heard a lot of older people say that the early 20s are the hardest part in anyones life, so i been trying to keep to the belief that itll get better on its own. who knows tho.. lol. i try to work out as much as possible but its all for hate and violence. just dont kno anymore.. hope the pills help
eskae Posts : 124
Join date : 20100503
Age : 29
Location : Vallejo, CA
Re: Rant Thread
Just don't commit suicide. Suicide is the most stupidest thing anyone can ever do.
I suggest getting some further education, or take up some sort of activity, like gym or a club. Get your mind off things.
The healthier your body is, the healthier your mental state is. That'll also reflect into your social life + could attract thehookers, lovely ladies.
Alternatively you can take up an instrument/paint/draw/write, become creative and get into an art scene of some sort.
Or just listen to liquid all day long
OR if you have the money, move to Cuba or London, for no reason whatsoever.
I suggest getting some further education, or take up some sort of activity, like gym or a club. Get your mind off things.
The healthier your body is, the healthier your mental state is. That'll also reflect into your social life + could attract the
Alternatively you can take up an instrument/paint/draw/write, become creative and get into an art scene of some sort.
Or just listen to liquid all day long
OR if you have the money, move to Cuba or London, for no reason whatsoever.
Menyth Posts : 56
Join date : 20100503
Age : 27
Location : Melbourne
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