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.io.*;
import java.util.*;
import java.lang.*;

public class Rextester{
    public static long modPow(long x,long y,long mod){
        long res=1;
        x=x%mod;
        while(y>0){
            if((y&1)==1){
                res=(res*x)%mod;
            }
            y=y>>1;
            x=(x*x)%mod;
        }
        return res;
    }
    public static void main(String[] args)throws IOException{
        long mod = 1000000007;
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        long x = new Long(st.nextToken());
        long n = new Long(st.nextToken());
        br.close();
        long ans=1;
        for(int i=2;i<=(long)Math.sqrt(x);i++){
            int count=0;
            while(x%i==0){
                x/=i;
                count++;
            }
            if(count>0){
                //System.out.print(i+" ");
                long y = n;
                long pow=0;
                while(y>0){
                    pow+=y/i;
                    y=y/i;
                }
                ans=((ans%mod)*(modPow(i,pow,mod))%mod)%mod;
            }
        }
        //System.out.println(ans);
        if(x>1){
            long y=n;
            long pow=0;
            while(y>0){
                pow+=y/x;
                y=y/x;
            }
            ans=((ans%mod)*(modPow(x,pow,mod))%mod)%mod;
        }
        System.out.println(ans);
    }
}          