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
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 ll poww(ll a,ll b){//pow会编译错误 23 ll ans=1%mod; 24 while(b){ 25 if(b&1) { 26 ans=ans*a%mod; 27 } 28 b>>=1; 29 a=a*a%mod; 30 } 31 return ans%mod; 32 } 33 int t; 34 ll n; 35 /* 36 i从1到n i*i*C(n,i)的累加和 37 38 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 39 两个司令可以是同一个人 40 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 41 2 : 两个人 A(n,2)*2^(n-2) 同理 ，2个人要考虑顺序 42 也就是 n*(n+1)*2^(n-2) 43 */ 44 45 int main() 46 { 47 scanf("%d",&t); 48 while(t--){ 49 scanf("%lld",&n); 50 if(n==1) {printf("1 ");continue;}//特判 51 ll ans=n*(n+1)%mod*poww(2,n-2)%mod; 52 printf("%lld ",ans); 53 } 54 return 0; 55 }