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 = 1000000007;
long long A(long long B, long long C) {
  if (C == 0) return 1;
  if (C % 2) return (A(B, C - 1) * B) % MOD;
  long long X = A(B, C / 2);
  return (X * X) % MOD;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  long long X, N;
  cin >> X >> N;
  set<long long> Y;
  for (int i = 0; i < (50); i++)
    for (int j = 0; j < (sqrt(X) + 1); j++) {
      if (j > 1)
        if (X % j == 0) {
          X /= j;
          Y.insert(j);
          break;
        }
    }
  Y.insert(X);
  auto itr = Y.begin();
  long long ans = 1;
  for (int i = 0; i < (Y.size()); i++) {
    long long T = (*itr);
    long long TT = T;
    long long cnt = 0;
    if (T > 1)
      while (TT <= N) {
        cnt += N / TT;
        if (N / TT >= T)
          TT *= T;
        else
          break;
      }
    itr++;
    ans = (ans * A(T, cnt)) % MOD;
  }
  cout << ans << endl;
}
