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 x, n, ans;
vector<long long> v;
void prime(long long x) {
  long long st = 2;
  long long x1 = x, x2 = sqrt(x) + 1;
  while (x) {
    if (x % st == 0) {
      if (st != x1) {
        v.push_back(st);
        if (x / st == st) {
          return;
        }
      }
      while (x % st == 0) {
        x /= st;
      }
    }
    ++st;
    if (st >= x2) break;
    if (st > x) break;
  }
  if (x > 1) {
    v.push_back(x);
  }
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> x >> n;
  prime(x);
  if (v.size() == 0) {
    v.push_back(x);
  }
  ans = 1;
  long long cnt;
  for (int i = 0; i < v.size(); i++) {
    long long temp = 1;
    for (int j = 1; j <= 100; j++) {
      if (temp >= n) {
        if (temp > n) temp /= v[i];
        break;
      }
      if (n / temp < v[i]) break;
      temp *= v[i];
    }
    cnt = 0;
    while (temp > 1) {
      ans = (ans % 1000000007 * power(temp, n / temp - cnt, 1000000007) %
             1000000007) %
            1000000007;
      cnt = n / temp;
      temp /= v[i];
    }
  }
  cout << ans << "\n";
}
