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 expo(long long int base, long long int exponent,
                   long long int mod) {
  long long int ans = 1;
  while (exponent != 0) {
    if (exponent & 1) ans = (1LL * ans * base) % mod;
    base = (1LL * base * base) % mod;
    exponent >>= 1;
  }
  return ans % mod;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) {
    long long int x, n;
    cin >> x >> n;
    long long int sx = sqrt(x) + 10;
    map<long long int, long long int> ma;
    for (long long int i = 2; i <= sx; i++) {
      while (x % i == 0) {
        x = x / i;
        ma[i]++;
      }
    }
    if (x != 1) ma[x]++;
    vector<long long int> v;
    for (auto i : ma) v.push_back(i.first);
    long long int ans = 1;
    for (auto i : v) {
      long long int p = i;
      long long int e = 0;
      long long int nn = n;
      for (long long int j = 1; true; j++) {
        e += nn / p;
        e = e % (1000000007 - 1);
        nn = nn / p;
        if (nn == 0) break;
      }
      long long int tmp = expo(i, e % (1000000007 - 1), 1000000007);
      ans = (ans * tmp) % 1000000007;
    }
    cout << ans;
  }
  return 0;
}
