#include <iostream>
using namespace std;
long long binaryPowerMethod(long long z,long long n, long long mod){
  long long ans = 1;
  while(n != 0){
    if(n%2){
      ans*=z;
      if(mod)ans%=mod;
    }
    n/=2;
    z*=z;
    if(mod)z%=mod;
  }
  return ans;
}
int main(){
    long long n,m;cin>>n>>m;
    cout << binaryPowerMethod(n,m,1000000007) << endl;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 <iostream>
using namespace std;
long long binaryPowerMethod(long long z,long long n, long long mod){
  long long ans = 1;
  while(n != 0){
    if(n%2){
      ans*=z;
      if(mod)ans%=mod;
    }
    n/=2;
    z*=z;
    if(mod)z%=mod;
  }
  return ans;
}
int main(){
    long long n,m;cin>>n>>m;
    cout << binaryPowerMethod(n,m,1000000007) << endl;
}

```