Teemo is very interested in convex polygon. There is a convex n-sides polygon, and Teemo connect every two points as diagonal lines, and he want to kown how many segments which be divided into intersections. Teemo ensure that any three diagonals do not intersect at a point.
As the result may be very large, please output the result mod 1000000007.
Input Format
The input contains several test cases, and the first line is a positive integer T indicating the number of test cases which is up to 100.
For each test case, the first line contains an integer n(3<=n<=10^18).
Output Format
For each test case, output a line containing an integer that indicates the answer.
样例输入
2 3 4
样例输出
0 4
1 #include <iostream> 2 #include <cstdio> 3 #include <algorithm> 4 #include <cstdlib> 5 #include <cstring> 6 #include <string> 7 #include <deque> 8 #include <map> 9 #include <vector> 10 #include <stack> 11 using namespace std; 12 #define ll long long 13 #define N 29 14 #define M 1000000000 15 #define gep(i,a,b) for(int i=a;i<=b;i++) 16 #define gepp(i,a,b) for(int i=a;i>=b;i--) 17 #define gep1(i,a,b) for(ll i=a;i<=b;i++) 18 #define gepp1(i,a,b) for(ll i=a;i>=b;i--) 19 #define mem(a,b) memset(a,b,sizeof(a)) 20 #define ph push_back 21 #define mod 1000000007 22 void egcd(ll a,ll b,ll &x,ll &y) 23 { 24 ll d=a; 25 if(!b) { 26 x=1;y=0; 27 return ; 28 } 29 else{ 30 egcd(b,a%b,y,x); 31 y-=(a/b)*x; 32 } 33 } 34 ll inv(int n){ 35 ll x,y; 36 egcd(n,mod,x,y); 37 return (x%mod+mod)%mod; 38 } 39 int t; 40 ll n; 41 /* 42 //任意三个无交点 43 n*(n-3) : 凸多边形的每个顶点与其他的点连接形成的线段数目 44 4*C(n,4) : 4个点形成形内1顶点，而1个点出现4个分割的线段 45 都出现了两次
(C(n,4)*4+n*(n-3))/2 46 / 47 int main() 48 { 49 scanf("%d",&t); 50 while(t--) 51 { 52 scanf("%lld",&n); 53 n%=mod;//避免溢出 54 ll ans=n*(n-3)%mod*(8+n*n%mod-3*n%mod+mod)%mod*inv(12)%mod; 55 printf("%lld ",ans); 56 } 57 return 0; 58 }