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>
using namespace std;
const int mod = (1e9) + 7;
bool isprime(long long int a) {
  if (a <= 1) return false;
  for (long long int i = 2; i < floor(sqrt(a) + 1); i++) {
    if (a % i == 0) {
      return false;
    }
  }
  return true;
}
long long int modpow(long long int x, long long int y) {
  long long int ans = 1;
  while (y > 0) {
    if (y & 1) {
      ans = ((ans % mod) * (x % mod)) % mod;
    }
    y = y >> 1;
    x = ((x % mod) * (x % mod)) % mod;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int x, n, ans = 1;
  cin >> x >> n;
  for (long long int i = 1; i < floor(sqrt(x)) + 1; i++) {
    if (x % i == 0 && isprime(i)) {
      long long int m = n / i;
      while (m) {
        ans = (ans * modpow(i, m)) % mod;
        m = m / i;
      }
    }
    if (x % i == 0 && isprime((x / i)) && i != (x / i)) {
      long long int m = n / (x / i);
      while (m) {
        ans = (ans * modpow((x / i), m)) % mod;
        m = m / (x / i);
      }
    }
  }
  cout << ans;
  return 0;
}
