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
#include <stdio.h>

long long int EasyPow(long long int x,long long int y,long long int z);

int main(){
 long long int m;
 long long int n;

 scanf("%lld %lld",&m,&n);

 printf("%lld\n",EasyPow(m,n,1000000007));

 return 0;
}

long long int EasyPow(long long int x,long long int y,long long int z){
 //if(y==0)return 1;
 long long int res=1;
 if(y>0){
 res=EasyPow(x,y/2,z); // ????????§?????°???y=0??????????????§
 
 //printf("[%u]\n",res);

 if(y%a==0)res=(res*res)%z;
 else res=(((res*res)%z)*x)%z; // ????????????1*x?????????????????¨?????????
 }
 return res;
 
}