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.io.*;
import java.util.*;

public class Main {

    public static void main(String[] args) throws IOException {
        new Main().solve();
    }
    static FastScanner in;
    static PrintWriter out;
    static long x,n;
    void read() throws IOException {
        in = new FastScanner(System.in);
      int t = 1;
      while (t-- > 0){
      x = in.nextLong();
      n = in.nextLong();
    }
    }
    static long mod = 1_000_000_007;
    long  pow(long a, long b) {
        if (b == 0)return 1;
        if (b % 2 == 1)return (pow(a, b - 1) * a) % mod;
        long t = pow(a, b / 2);
        return (t * t) % mod;
    }

    long pow1(long a, long b) {
        if (b == 0)return 1;
        if (b % 2 == 1)return (pow1(a, b - 1) * a);
        long t = pow1(a, b / 2);
        return t * t;
    }

    long canMax(long n , long div){
        long max = 0;
        while (n >= div) {
            n/=div;
            max++;
        }
        return max;
    }
    void solve() throws IOException {
        read();
        TreeSet<Long> div = new TreeSet<>();
        for (int i = 1; i <= Math.sqrt(x); i++) {
            if (x % i == 0) {
                if (isPrime(i)) div.add((long) i);
                if (isPrime(x / i)) div.add(x / i);
            }
        }
        long ans = 1;
        while (!div.isEmpty()) {
            long d = div.pollFirst();
            long maxIn = canMax(n, d);
            long num = pow1(d, maxIn);
            long tmp = 0;
            while (maxIn != 0) {
                ans = (ans * pow(num % mod, n / num - tmp)) % mod;
                tmp = n / num;
                maxIn--;
                num = pow1(d, maxIn);
            }
        }
        System.out.println(ans);
    }
    boolean isPrime(long q){
         if(q == 1)return false;
        for (int i = 2; i <= Math.sqrt(q); i++) {
            if(q % i == 0)return false;
        }
        return true;
    }
}
class FastScanner {
    BufferedReader br;
    StringTokenizer st = new StringTokenizer("");

    FastScanner(InputStream f) throws FileNotFoundException {
        br = new BufferedReader(new InputStreamReader(f));
    }

    String next() throws IOException {
        while (!st.hasMoreTokens()) {
            st = new StringTokenizer(br.readLine());
        }
        return st.nextToken();
    }

    long nextLong() throws IOException {
        return Long.parseLong(next());
    }
}