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 m(long long a) {
  return ((a % 1000000007) + 1000000007) % 1000000007;
}
set<long long> s;
void pf(long long n) {
  while (n % 2 == 0) {
    s.insert(2);
    n /= 2;
  }
  for (long long i = 3; i * i <= n; i += 2) {
    while (n % i == 0) {
      s.insert(i);
      n /= i;
    }
  }
  if (n > 2) s.insert(n);
}
long long p(long long a, long long po) {
  if (po == 0) return 1ll;
  long long x = p(a, po / 2);
  if (po & 1) {
    return m(m(x * x) * a);
  }
  return m(x * x);
}
int main() {
  long long x, n;
  cin >> x >> n;
  pf(x);
  long long ans = 1;
  vector<long long> freq(150, 0);
  for (long long i : s) {
    int ind = 0;
    long long k = i;
    while (n >= k) {
      freq[ind++] = n / k;
      if (n / k < i) break;
      k *= i;
    }
    for (int j = ind - 1; j > 0; j--) {
      for (int l = 0; l < j; l++) {
        freq[l] -= freq[j];
      }
    }
    long long k1 = i;
    for (int j = 0; j < ind; j++) {
      assert(freq[j] >= 0);
      long long temp = p(k1, freq[j]);
      ans = m(ans * temp);
      freq[j] = 0;
      k1 *= i;
      k1 = m(k1);
    }
  }
  cout << ans << endl;
}
