Easy Summation

You are encountered with a traditional problem concerning the sums of powers. 
Given two integers nn and kk. Let f(i)=ikf(i)=ik, please evaluate the sum f(1)+f(2)+...+f(n)f(1)+f(2)+...+f(n). The problem is simple as it looks, apart from the value of nnin this question is quite large. 
Can you figure the answer out? Since the answer may be too large, please output the answer modulo 109+7109+7.

InputThe first line of the input contains an integer T(1≤T≤20)T(1≤T≤20), denoting the number of test cases. 
Each of the following TT lines contains two integers n(1≤n≤10000)n(1≤n≤10000) and k(0≤k≤5)k(0≤k≤5). 
OutputFor each test case, print a single line containing an integer modulo 109+7109+7.Sample Input

3
2 5
4 2
4 1

Sample Output

33
30
10

题意：例第一组就是1的5次方加2的5次方（用快速幂）

#include <iostream>
#include<algorithm>
#include<stdlib.h>
#include<stdio.h>
#include<string.h>
#include<math.h>
using namespace std;
long long int f(long long int n,long long int k)
{
    long long int ans=1;
    while(k!=0)
    {
        if(k&1)
            ans=(n%1000000007)*(ans%1000000007);
        n=((n%1000000007)*(n%1000000007))%1000000007;
        k>>=1;
    }


    return ans;
}
int main()
{
    long long int T;
    scanf("%lld",&T);
    while(T--)
    {
        long long int n,k,i,sum=0;
        scanf("%lld%lld",&n,&k);
        for(i=1;i<=n;i++)
        {
            sum+=f(i,k)%1000000007;
        }
        printf("%lld
",sum%1000000007);
    }
    return 0;
}