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.
#include <bits/stdc++.h>
int primes(long long x, int* primelist) {
  int n = 0;
  for (int i = 2; i <= sqrt(x) + 1; i++) {
    if (x % i == 0) {
      primelist[n] = i;
      n++;
      while (x % i == 0) {
        x /= i;
      }
    }
  }
  if (x != 1 && x != 0) {
    primelist[n] = x;
    n++;
  }
  return n;
}
long long func(int* primelist, int listlen, long long n) {
  long long ans = 1;
  for (int i = 0; i < listlen; i++) {
    long long p = primelist[i];
    long long s = 0;
    long long x = n;
    while (x) {
      s += x / p;
      x /= p;
    }
    while (s) {
      if (s & 1) {
        ans = ans * p % 1000000007;
      }
      p = p * p % 1000000007;
      s >>= 1;
    }
  }
  return ans;
}
int main() {
  long long X, N;
  scanf("%lld%lld", &X, &N);
  int p[32768];
  int n;
  long long ans;
  n = primes(X, p);
  ans = func(p, n, N);
  printf("%lld\n", ans);
  return 0;
}
