Problem Description

RXD is a good mathematician. 
One day he wants to calculate:

output the answer module 109+7. 
1≤n,k≤1018 

p1,p2,p3…pk are different prime numbers

Input

There are several test cases, please keep reading until EOF. 
There are exact 10000 cases. 
For each test case, there are 2 numbers n,k.

Output

For each test case, output “Case #x: y”, which means the test case number and the answer.

Sample Input

10 10

Sample Output

Case #1: 999999937

题目大意：

C++

#include<bits/stdc++.h>
#define mod 1000000007
using namespace std;
long long pow(long long a,long long b)
{
   long long int ans = 1,base = a;
    while(b!=0)
    {
        if(b&1){
            base%=(long long)mod;
            ans%=(long long)mod;
            ans *= base;
            ans%=(long long)mod;
        }
        base%=(long long)mod;
        base *= base;
        base%=(long long)mod;
        b>>=1;
    }
    return ans;
}
int main()
{
    long long int n,k;
    int t=0;
    while(~scanf("%lld %lld",&n,&k)){
            t++;
        printf("Case #%d: ",t);
        printf("%lld\n",pow(n,k));
    }
    return 0;
}