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;
char x[11][11];
long long int arra[200009];
map<long long int, long long int> ans;
vector<long long int> code;
map<long long int, long long int> subseq;
map<long long int, long long int> visited;
map<char, long long int> zerone;
long long int Mp[1009][1009];
bool prime_check[100000];
long long int mod = 1000000007;
bool prime_chcek(long long int n) {
  long long int m, i, j, k, l, p, q, r, s, t;
  p = sqrt(n);
  for (i = 2; i <= p; i++) {
    if (n % i == 0) return 1;
  }
  return 0;
}
long long int bigmod(long long int a, long long int p, long long int m) {
  long long res = 1;
  long long x = a;
  while (p) {
    if (p & 1) {
      res = (res * x) % m;
    }
    x = (x * x) % m;
    p = p >> 1;
  }
  return res;
}
long long int sum(long long int p, long long int n) {
  long long int q = 0, r, s, t;
  while (n) {
    q = q + n / p;
    n = n / p;
  }
  return bigmod(p, q, mod) % mod;
}
int main() {
  long long int i, j, k, l, p = 1, q = 1000000007, r = 1, s, t, m, n, two = 2,
                            x;
  bool pq;
  cin >> x >> n;
  pq = prime_chcek(x);
  k = sqrt(x);
  for (i = 2; i <= k; i++) {
    if (x % i == 0) {
      r = r * sum(i, n) % mod;
      while (x % i == 0) {
        x = x / i;
      }
    }
  }
  if (x != 1) {
    r = r * sum(x, n) % mod;
  }
  cout << r << '\n';
}
