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.Scanner;

public class cf1228C {
    private static boolean[] getPrime(int n) {
        boolean[] res = new boolean[n];
        res[0] = res[1] = true;
        for(int i = 2; i < n; i++) {
            if(!res[i]) {
                for(int j = i * 2; j < n; j += i) {
                    res[j] = true;
                }
            }
        }
        return res;
    }
    private static final long MOD = (int)1e9 + 7;
    private static long qpow(int a, long x) {
        long res = 1, cur = a;
        while(x != 0) {
            if((x & 1) == 1) {
                res *= cur;
                res %= MOD;
            }
            cur *= cur;
            cur %= MOD;
            x >>= 1;
        }
        return res;
    }
    private static long f(int p, long n) {
        long cnt = 0;
        long cur = 1;
        do {
            cur *= p;
            cnt += n / cur;
        } while(cur <= n / p);
        return qpow(p, cnt);
    }
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int x = in.nextInt();
        long n = in.nextLong();

        boolean notPrime[] = getPrime((int)Math.sqrt(x) + 5);
        long ans = 1;
        for(int i = 2; i < notPrime.length; i++) {
            if(!notPrime[i] && x % i == 0) {
                ans *= f(i, n);
                ans %= MOD;
                while(x % i == 0) {
                    x /= i;
                }
            }
        }
        if(x > 1) {
            ans *= f(x, n);
            ans %= MOD;
        }
        System.out.println(ans);
    }
}
