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;
int md = 1e9 + 7;
long long int mod_exp(long long int x, long long int n) {
  long long int res = 1;
  while (n != 0) {
    if (n & 1) {
      res = (res * x) % md;
    }
    n = n >> 1;
    x = (x * x) % md;
  }
  return res;
}
int main() {
  int i = 2;
  long long int x, n, ans = 1;
  cin >> x >> n;
  vector<int> pfs;
  if (x % 2 == 0) {
    pfs.push_back(2);
    while (x % 2 == 0) {
      x = x / 2;
    }
  }
  int rtx = sqrt(x);
  for (int i = 3; i < rtx + 1; i += 2) {
    if (x % i == 0) {
      pfs.push_back(i);
      while (x % i == 0) {
        x = x / i;
      }
    }
  }
  if (x > 2) {
    pfs.push_back(x);
  }
  vector<long long int> cnt;
  int pfs_size = pfs.size();
  for (int i = 0; i < pfs_size; i++) {
    long long int dummy = n;
    cnt.push_back(0);
    while (n != 0) {
      cnt[i] += (n / (long long int)pfs[i]);
      n = (n / (long long int)pfs[i]);
    }
    n = dummy;
  }
  for (int i = 0; i < pfs_size; i++) {
    ans = (ans * (mod_exp((long long int)pfs[i], (long long int)cnt[i]))) % md;
  }
  cout << ans << endl;
  return 0;
}
