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;
bool OF(long long a, long long b) {
  if ((a * b) / a == b)
    return false;
  else
    return true;
}
long long mo = 1e9 + 7;
vector<long long> v;
long long ans = 1, now = 1, x, n;
int main() {
  ios_base::sync_with_stdio(0);
  long long d;
  cin >> x >> n;
  set<long long> se;
  d = x;
  for (int i = 2; i * i <= x; i++) {
    if (d % i == 0ll) {
      se.insert(i);
      d /= i;
      i--;
    }
  }
  for (auto i : se) {
    v.push_back(i);
  }
  if (d != 1) v.push_back(d);
  for (long long i : v) {
    long long d = i, sum = 0;
    while (d <= n) {
      long long k = n / d;
      sum += k;
      if (!OF(d, i))
        d *= i;
      else
        break;
    }
    long long r = i;
    for (int j = 0; j < 63; j++) {
      if (sum & (1ll << j)) {
        ans *= r;
        ans %= mo;
      }
      r = (r * r);
      r %= mo;
    }
  }
  cout << ans << endl;
}
