This is a small but ancient game. You are supposed to write down the numbers 1,2,3,...,2n - 1,2n consecutively in clockwise order on the ground to form a circle,and then,to draw some straight line segments to connect them into number pairs. Every number must be connected to exactly one another. And,no two segments are allowed to intersect.

It's still a simple game,isn't it? But after you've written down the 2n numbers,can you tell me in how many different ways can you connect the numbers into pairs? Life is harder,right?

Each line of the input file will be a single positive number n,except the last line,which is a number -1. You may assume that 1 <= n <= 100.

For each n,print in a single line the number of ways to connect the 2n numbers into pairs.

  
  

   
   2
3
-1
  
  

  
  

   
   2
5
  
  

Asia 2004,Shanghai (Mainland China),Preliminary?

题目大意：2n个连线，不能相交，求最多有多少种可能的握手方法？

考察知识点：大数，卡特兰数

//考察知识点：卡特兰数　＆＆　大数。
/*
h(1)=1,h(0)=1;h(n)=((4*n-2)/(n+1))*h(n-1);(其中ｎ＞＝２) 
*/ 
#include<stdio.h>
#include<string.h>
#define inf 110
int a[inf][inf];
int i,j,temp,len,n;
void f()
{
	temp=0;len=1;
	memset(a,sizeof(a));
	a[1][0]=1;
	for(i=2;i<=100;++i)
	{
		for(j=0;j<len;++j)
		{
			a[i][j]=a[i-1][j]*(4*i-2);
		}
		for(j=0;j<len;++j)
		{
			temp+=a[i][j];
			a[i][j]=temp%10;
			temp/=10;
		}
		while(temp)
		{
			a[i][len++]=temp%10;
			temp/=10;
		}
		//len--;
		for(j=len-1;j>=0;--j)
		{
			temp=temp*10+a[i][j];
			a[i][j]=temp/(i+1);
			temp%=i+1;
		}
		while(!a[i][len-1])
		len--;
	}
}
int main()
{
	f();
	while(~scanf("%d",&n),n!=-1)
	{
		for(i=inf-1;i>=0&&!a[n][i];--i);
		for(;i>=0;--i)
		printf("%d",a[n][i]);
		puts("");
	}
	return 0;
}