Let's introduce some definitions that will be needed later.

Let prime(x) be the set of prime divisors of x. For example, prime(140) = \{ 2, 5, 7 \}, prime(169) = \{ 13 \}.

Let g(x, p) be the maximum possible integer p^k where k is an integer such that x is divisible by p^k. For example:

  * g(45, 3) = 9 (45 is divisible by 3^2=9 but not divisible by 3^3=27), 
  * g(63, 7) = 7 (63 is divisible by 7^1=7 but not divisible by 7^2=49). 



Let f(x, y) be the product of g(y, p) for all p in prime(x). For example:

  * f(30, 70) = g(70, 2) ⋅ g(70, 3) ⋅ g(70, 5) = 2^1 ⋅ 3^0 ⋅ 5^1 = 10, 
  * f(525, 63) = g(63, 3) ⋅ g(63, 5) ⋅ g(63, 7) = 3^2 ⋅ 5^0 ⋅ 7^1 = 63. 



You have integers x and n. Calculate f(x, 1) ⋅ f(x, 2) ⋅ … ⋅ f(x, n) mod{(10^{9} + 7)}.

Input

The only line contains integers x and n (2 ≤ x ≤ 10^{9}, 1 ≤ n ≤ 10^{18}) — the numbers used in formula.

Output

Print the answer.

Examples

Input


10 2


Output


2


Input


20190929 1605


Output


363165664


Input


947 987654321987654321


Output


593574252

Note

In the first example, f(10, 1) = g(1, 2) ⋅ g(1, 5) = 1, f(10, 2) = g(2, 2) ⋅ g(2, 5) = 2.

In the second example, actual value of formula is approximately 1.597 ⋅ 10^{171}. Make sure you print the answer modulo (10^{9} + 7).

In the third example, be careful about overflow issue.
import java.math.BigInteger;
import java.util.Scanner;

public class Main {
    static long[] v=new long[1005];
    static int cnt=0;
    static long mod=(long)1e9+7;
    static boolean prime(long n){
        if(n==1)
            return false;
        for(long i=2;i*i<=n;i++){
            if(n%i==0)
                return false;
        }
        return true;
    }
    static void solve(long n){
        for(long i=1;i*i<=n;i++){
            if(n%i==0){
                if(prime(i))
                    v[cnt++]=i;
                if(i!=n/i && prime(n/i))
                    v[cnt++]=n/i;
            }
        }
    }
    static long fpow(long a,BigInteger n){
        long res=1,base=a;
        while(n.compareTo(BigInteger.valueOf(0))!=0){
            if(n.mod(BigInteger.valueOf(2)).compareTo(BigInteger.valueOf(1))==0)
                res=res*base%mod;
            base=base*base%mod;
            n=n.divide(BigInteger.valueOf(2));
        }
        return res;
    }
    public static void main(String args[]){
        Scanner sc=new Scanner(System.in);
        long x=sc.nextLong(),n=sc.nextLong();
        solve(x);
        long res=1;
        for(int i=0;i<cnt;i++){
            BigInteger base=BigInteger.valueOf(v[i]);
            while(base.compareTo(BigInteger.valueOf(n))<=0){
                res=res*fpow(v[i],BigInteger.valueOf(n).divide(base))%mod;
                base=base.multiply(BigInteger.valueOf(v[i]));
            }
        }
        System.out.println(res);
    }
}
