Spoj-TRNGL Make Triangle

 Make Triangle

Chayanika loves Mathematics. She is learning a new chapter geometry. While reading the chapter a question came in her mind. Given a convex polygon of n sides. In how many ways she can break it into triangles, by cutting it with (n-3) non-adjacent diagonals and the diagonals do not intersect.

 

Input

First line of the input will be an integer t (1<=t<=100000) which is the no of test cases. Each test case contains a single integer n (3<=n<=1000) which is the size of the polygon.

Output

For each test case output the no of ways %100007.

Example

Input:
2
3
5

 Output:
1
5

很迷……答案显然就是卡特兰数
然后在计算的时候出现了一些小小的偏差
c[i]=c[i-1]*(4i-2)/(i+1)的递推式是不行的，因为特么取模的1e5+7不是质数，i+1的逆元怎么搞啊……
然后用n^2的c[k]*c[i-k]的递推了

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<ctime>
#define LL long long
#define inf 0x7ffffff
#define pa pair<int,int>
#define mkp(a,b) make_pair(a,b)
#define pi 3.1415926535897932384626433832795028841971
#define mod 100007
using namespace std;
inline LL read()
{
    LL x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
LL catlan[1010];
int main()
{
    catlan[1]=catlan[0]=1;
    for (int i=2;i<=1000;i++)
        for (int j=0;j<i;j++)
            catlan[i]=(catlan[i]+catlan[j]*catlan[i-j])%mod;
    int T=read();
    while (T--)printf("%lld\n",catlan[read()-1]);
}