#include<bits/stdc++.h>
using namespace std;
const int mod=1e9+7;
long long modpow(long long x,long long y){
    if(y==1)return x%mod;
    else if(y%2) return (x*modpow(x,y-1))%mod;
    long long t=modpow(x,y/2);
    return t*t%mod;
}
int main(){
    long long m,n;scanf("%lld %lld",&m,&n);
    printf("%lld\n",modpow(m,n));
}

### Prompt

Develop a solution in cpp to the problem described below:

For given integers m and n, compute mn (mod 1,000,000,007). Here, A (mod M) is the remainder when A is divided by M.

Constraints

* 1 ≤ m ≤ 100
* 1 ≤ n ≤ 109

Input


m n


Two integers m and n are given in a line.

Output

Print mn (mod 1,000,000,007) in a line.

Examples

Input

2 3


Output

8


Input

5 8


Output

390625

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
const int mod=1e9+7;
long long modpow(long long x,long long y){
    if(y==1)return x%mod;
    else if(y%2) return (x*modpow(x,y-1))%mod;
    long long t=modpow(x,y/2);
    return t*t%mod;
}
int main(){
    long long m,n;scanf("%lld %lld",&m,&n);
    printf("%lld\n",modpow(m,n));
}

```