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 lint = long long;
const size_t max_primer = 100000;
const lint mod = 1e9 + 7;
bool is_not_primer[max_primer];
std::vector<int> primer;
lint quickPow(const lint& base, lint pow) {
  lint temp = base, res = 1;
  while (pow) {
    if (pow & 1) {
      res = (res * temp) % mod;
    }
    temp = (temp * temp) % mod;
    pow >>= 1;
  }
  return res;
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::cout.tie(nullptr);
  for (int i = 2; i < max_primer; i++) {
    if (!is_not_primer[i]) {
      primer.push_back(i);
      for (int j = 2; i * j < max_primer; j++) {
        is_not_primer[i * j] = true;
      }
    }
  }
  lint x, n;
  std::cin >> x >> n;
  lint res = 1;
  for (const auto& p : primer) {
    if ((lint)p * p > x) {
      break;
    }
    if (x % p != 0) {
      continue;
    }
    while (x % p == 0) {
      x /= p;
    }
    lint pow = 0, temp = n;
    while (temp / p != 0) {
      pow += (temp / p);
      temp /= p;
    }
    res = (res * quickPow(p, pow)) % mod;
  }
  if (x != 1) {
    lint pow = 0, temp = n;
    while (temp / x != 0) {
      pow += (temp / x);
      temp /= x;
    }
    res = (res * quickPow(x, pow)) % mod;
  }
  std::cout << res << std::endl;
  return 0;
}
