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=,f=;char ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
     return x*f;
 }
 LL catlan[];
 int main()
 {
     catlan[]=catlan[]=;
     for (int i=;i<=;i++)
         for (int j=;j<i;j++)
             catlan[i]=(catlan[i]+catlan[j]*catlan[i-j])%mod;
     int T=read();
     while (T--)printf("%lld\n",catlan[read()-]);
 }

Spoj TRNGL