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;
long long int binary_exp(long long int a, long long int n) {
  long long int i = 1;
  while (n > 0) {
    if (n % 2 == 1) i = (i * a) % 1000000007;
    a = (a * a) % 1000000007;
    n /= 2;
  }
  return i;
}
long long int IsPrime(long long int n) {
  if (n == 0 || n == 1) return n;
  if (n % 2 == 0) return (long long int)2;
  if (n % 3 == 0)
    return (long long int)3;
  else {
    for (long long int i = 5; i * i <= n; i += 6) {
      if (n % i == 0) return i;
      if (n % (i + 2) == 0) return (i + 2);
    }
    return n;
  }
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  srand(time(NULL));
  ;
  long long int n, i, j, k, a, b, x, ans = 1;
  cin >> x >> n;
  vector<long long int> prime;
  k = 0;
  while (x % 2 == 0) {
    k++;
    x /= 2;
  }
  if (k > 0) prime.push_back(2);
  for (i = 3; i <= sqrt(x); i += 2) {
    k = 0;
    while (x % i == 0) {
      k++;
      x /= i;
    }
    if (k > 0) prime.push_back(i);
  }
  if (x > 2) prime.push_back(x);
  for (i = 0; i < prime.size(); i++) {
    k = 0;
    j = n;
    while (j > 0) {
      k += (j / prime[i]);
      j /= prime[i];
    }
    ans = (ans * binary_exp(prime[i], k)) % 1000000007;
  }
  cout << ans;
}
