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 unsigned x, n, csum[100], ans = 1;
vector<long long unsigned> v;
long long unsigned power(long long unsigned b, long long unsigned p) {
  if (p == 0) return 1;
  if (p == 1) return b % 1000000007;
  long long unsigned x = power(b, p / 2) % 1000000007;
  x = (x * x) % 1000000007;
  if (p & 1ULL) x = (x * b) % 1000000007;
  return x;
}
int main() {
  cin >> x >> n;
  long long unsigned fc = x;
  for (long long unsigned i = 2; i * i <= fc; i++) {
    if (fc % i == 0) {
      v.push_back(i);
      while (fc % i == 0) fc = fc / i;
    }
  }
  if (fc != 1) v.push_back(fc);
  for (int i = 0; i < v.size(); i++) {
    long long unsigned p = v[i];
    long long unsigned p1 = p;
    long long unsigned n1 = n;
    while (p <= n1) {
      ans = (ans * power(p, n1 / p)) % 1000000007;
      n1 = n1 / p;
    }
  }
  cout << ans << endl;
  return 0;
}
