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.
// practice with kaiboy
import java.io.*;
import java.util.*;

public class CF1228C extends PrintWriter {
	CF1228C() { super(System.out, true); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF1228C o = new CF1228C(); o.main(); o.flush();
	}

	static final int MD = 1000000007;
	long solve(long n, int p) {
		return n == 0 ? 0 : solve(n / p, p) + n / p;
	}
	long power(int a, long k) {
		if (k == 0)
			return 1;
		long p = power(a, k >> 1);
		p = p * p % MD;
		if ((k & 1) == 1)
			p = p * a % MD;
		return p;
	}
	void main() {
		int x = sc.nextInt();
		long n = sc.nextLong();
		long ans = 1;
		for (int p = 2; p <= x / p; p++)
			if (x % p == 0) {
				ans = ans * power(p, solve(n, p)) % MD;
				while (x % p == 0)
					x /= p;
			}
		if (x > 1) {
			int p = x;
			ans = ans * power(p, solve(n, p)) % MD;
		}
		println(ans);
	}
}
