#include <iostream>
using ll = long long;
#define rep(i, n) for(ll i = 0; i < n; i++)
using namespace std;

ll modpow(ll a, ll n, ll mod){
  ll res = 1;
  while(n > 0){
    if (n & 1) res = res * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return res;
}

int main(){
  ll n, m;
  cin >> n >> m;
  cout << modpow(n, m, 1000000007) << endl;
  
  return 0;
}

### 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 <iostream>
using ll = long long;
#define rep(i, n) for(ll i = 0; i < n; i++)
using namespace std;

ll modpow(ll a, ll n, ll mod){
  ll res = 1;
  while(n > 0){
    if (n & 1) res = res * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return res;
}

int main(){
  ll n, m;
  cin >> n >> m;
  cout << modpow(n, m, 1000000007) << endl;
  
  return 0;
}

```