ACM训练联盟周赛 Teemo's formula

Teemo has a formula and he want to calculate it quickly.

The formula is .

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 10^5.

For each test case, the first line contains an integer n(1<=n<=10^9).

Output Format

For each test case, output a line containing an integer that indicates the answer.

样例输入

2
2
3

样例输出

6
24

 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cstdlib>
 #include <cstring>
 #include <string>
 #include <deque>
 #include <map>
 #include <vector>
 #include <stack>
 using namespace std;
 #define  ll long long
 #define  N   29
 #define  M 1000000000
 #define  gep(i,a,b)  for(int  i=a;i<=b;i++)
 #define  gepp(i,a,b) for(int  i=a;i>=b;i--)
 #define  gep1(i,a,b)  for(ll i=a;i<=b;i++)
 #define  gepp1(i,a,b) for(ll i=a;i>=b;i--)
 #define  mem(a,b)  memset(a,b,sizeof(a))
 #define  ph  push_back
 #define  mod  1000000007
 ll poww(ll a,ll b){//pow会编译错误
     ll ans=%mod;
     while(b){
         if(b&)  {
             ans=ans*a%mod;
         }
         b>>=;
         a=a*a%mod;
     }
     return ans%mod;
 }
 int t;
 ll n;
 /*
 i从1到n  i*i*C(n,i)的累加和

 在N个人里面选若干人，再选一个正司令、副司令 的方法数目
 两个司令可以是同一个人
 1 ： 同一人 C(n,1)*2^(n-1)//一定至少有一个人是司令，其他的N-1个人（每人有两种可能性）可以是，也可以不是
 2 :  两个人 A(n,2)*2^(n-2) 同理 ，2个人要考虑顺序
 也就是 n*(n+1)*2^(n-2)
 */

 int main()
 {
     scanf("%d",&t);
     while(t--){
         scanf("%lld",&n);
         if(n==) {printf("1\n");continue;}//特判
         ll ans=n*(n+)%mod*poww(,n-)%mod;
         printf("%lld\n",ans);
     }
     return ;
 }