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 PrimesAndMultilication {
	static int mod = (int)1e9+7;
	public static void main(String[] args) {
		Scanner s = new Scanner(System.in);
		int x = s.nextInt();
		long n = s.nextLong();
		long ans = 1;
		Set<Integer> factors= factorsFun(x);

		
				for(int factor:factors) {
				ans = (ans%mod * fun(factor,n)%mod)%mod;
				}
			
			System.out.println(ans);
		

	}

	 static long fun(int p, long n) {
		 long count = 0;
		 while (n >= p) {
		 count += n / p;
		 n /= p;
		 }
	return expo(p, count);
		}
	private static long expo(long n, long k) {
		if(k==0)
			return 1L;
		if(k==1) return n;
		
		if(k%2==0) {
			long y = expo(n, k/2)%mod;
			return (y*y)%mod;
		}
		return ((n%mod)*(expo(n, k-1))%mod)%mod;
	}
	private static Set<Integer> factorsFun(int x) {
		Set<Integer> set = new HashSet<>();
		for(int i=2;i*i<=x;i++) {
			
			while(x%i==0) {
				x/=i;
				set.add(i);
			}
		}
		
		if(x>1) {
			set.add(x);
		}
		return set;
	}

}
