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.ArrayList;
import java.util.Scanner;
import java.math.BigInteger;

public class Main {
    static final BigInteger MOD = BigInteger.valueOf(1000000007);
    static boolean isPrime(int num) {
        if (num == 1) return false;
        if (num == 2) return true;
        if (num % 2 == 0) return false;
        int sqr = (int) Math.sqrt(num);
        for (int i = 3; i <= sqr; i += 2)
            if (num % i == 0)
                return false;
        return true;
    }

    static ArrayList<Integer> primes(int n) {
        ArrayList<Integer> primes = new ArrayList<>();
        for (int i = 1; i * i <= n; i++) {
            if (n % i == 0 && isPrime(i))
                primes.add(i);
            if (n / i * i == n && isPrime(n / i) && (n/i != i))
                primes.add(n / i);
        }
        return primes;
    }

    static BigInteger compute(int NN, long x) {
        BigInteger n = BigInteger.valueOf(NN);
        BigInteger org = BigInteger.valueOf(NN);
        BigInteger res = BigInteger.ONE;
        while (n.compareTo(BigInteger.valueOf(x)) <= 0) {
            BigInteger exp = (BigInteger.valueOf(x).divide(n)).subtract(BigInteger.valueOf(x).divide(n.multiply(org)));
//            res = res.mod(MOD);
            res = (res.mod(MOD).multiply(n.modPow(exp, MOD)).mod(MOD)).mod(MOD);
//            res = res.mod(MOD);
            n = n.multiply(org);
        }
        return res.mod(MOD);
    }
    static void solve(int x, long n) {
        ArrayList<Integer> prime = primes(x);
        BigInteger ans = BigInteger.ONE;
        for (int a : prime)
            ans = ans.multiply(compute(a, n));
        System.out.println(ans.mod(MOD));
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
            int x = sc.nextInt();
            long n = sc.nextLong();
            solve(x, n);
        
    }
}