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 fexp(long long int a, long long int b) {
  b %= (1000000007 - 1);
  if (b == 0) return 1;
  a %= 1000000007;
  long long int small = fexp(a, b / 2);
  small %= 1000000007;
  small *= small;
  small %= 1000000007;
  if (b & 1) small *= a;
  small %= 1000000007;
  return small;
}
int main() {
  unsigned long long int X, x, n, ans = 1ull, N;
  cin >> x >> n;
  X = x;
  set<int> primes;
  if (x > 0 and !(x & 1)) {
    primes.insert(2);
    while (!(x & 1) and x > 0) x >>= 1;
  }
  int i = 3;
  while (i * i <= X) {
    if (!(x % i)) {
      primes.insert(i);
      while (x and !(x % i)) x /= i;
    }
    i += 2;
  }
  if (x > 1) primes.insert(x);
  x = X;
  N = n;
  long long int powerk, temp;
  for (auto a : primes) {
    powerk = a;
    temp = 0;
    n = N;
    while ((n / powerk)) temp += (n / powerk), n /= a;
    ans *= (fexp(a, temp) % 1000000007);
    ans %= 1000000007;
  }
  cout << ans << endl;
  return 0;
}
