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;
void factors(long long n, vector<long long>& a) {
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      a.push_back(i);
      while (n % i == 0) n /= i;
    }
  }
  if (n > 1) a.push_back(n);
}
long long expn(long long n, long long p) {
  long long exp = 0;
  long long tmp = p;
  while (n > 0) {
    exp += n / p;
    n /= p;
  }
  return exp;
}
long long power_mod(long long n, long long pwr) {
  long long res = 1;
  while (pwr) {
    if (pwr % 2 == 1) res = (res * n) % 1000000007;
    n = (n * n) % 1000000007;
    pwr >>= 1;
  }
  return res;
}
void Solve() {
  long long x, n;
  cin >> x >> n;
  vector<long long> prime;
  factors(x, prime);
  long long ans = 1;
  for (int i = 0; i < prime.size(); i++) {
    long long exp = expn(n, prime[i]);
    ans *= power_mod(prime[i], exp);
    ans %= 1000000007;
  }
  cout << ans << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    Solve();
  }
  return 0;
}
