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;
vector<long long int> v;
long long int mod = 1000000007;
long long int bigmod(long long int a, long long int p, long long int m) {
  if (p == 0) return 1;
  if (p & 1) {
    return ((a % m) * (bigmod(a, p - 1, m))) % m;
  } else {
    long long int tmp = bigmod(a, p / 2, m);
    return (tmp * tmp) % m;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  unsigned long long int n, m, t, l = 2, r, ans = 1, p, u, flag, x, f = 0, e;
  long long int i;
  cin >> n >> m;
  t = sqrt(n);
  while (l <= t) {
    if (n % l == 0) {
      while (n % l == 0) {
        n /= l;
      }
      v.push_back(l);
    }
    l++;
  }
  if (n > 1) {
    v.push_back(n);
  }
  for (i = 0; i < v.size(); i++) {
    r = v[i];
    p = 1;
    x = 1;
    while (p <= m) {
      p *= r;
      if (p / r != x) {
        p = x;
        f = 1;
        break;
      }
      x = p;
    }
    if (f == 0) {
      p /= r;
    }
    flag = 0;
    while (p >= 1) {
      if (flag == 0) {
        flag = 1;
        u = (m / p);
        u = bigmod(p, u, mod);
        ans = ((ans % mod) * (u % mod)) % mod;
        p /= r;
        continue;
      }
      e = r * p;
      u = ((m / p) - (m / e));
      u = bigmod(p, u, mod);
      ans = ((ans % mod) * (u % mod)) % mod;
      p /= r;
    }
  }
  cout << ans << "\n";
}
