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.util.*;

public class prime{
	
	public static void main(String args[])
	{ 
		Scanner in = new Scanner(System.in);

		long x = in.nextLong();
		long n = in.nextLong();
       
        
        long val =1;
        long yo = (long)Math.pow(x,0.5)+1;
        long i =2;
        long z=x;
        boolean check = true;
        while(i<=yo)
        {
        	if(z%i==0)
        	{
        	 
        	 val = (val*pro(n,i))%1000000007;	
        	}
        	while(z%i==0)
        	{
        	 z=z/i;	
        	}
        
        	i++;
        }

        if(z!=1)
        {
          
          val = (val*pro(n,z))%1000000007; 	
        }
        System.out.println(val);


		
	}
	public static long power(long a,long p)
	{
		if(p==0)
		return 1;
		long val=power(a,p/2)%1000000007;
		if(p%2==0)
		return (val*val)%1000000007;
		else
		return (((val*val)%1000000007)*a)%1000000007;
	}
	public static long pro(long n, long p)
	{
		if(n<p)
		return 1;
		long val =1;
		long z=n;
        long q=p;

        while(q<=z)
        {
        	if(q<=z/p)
        	q=q*p;
        	else
        	break;
        }
        
       
       
        while(q>1)
        {
        	
        	long count = n/q;
        	val =(val* power(p,count))%1000000007;
        	q=q/p;
        }

		return val;
	}
}