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;
template <typename T>
T gcd(T a, T b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
template <typename T>
T pow(T a, T b, long long m) {
  T ans = 1;
  while (b > 0) {
    if (b % 2 == 1) ans = (ans * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return ans % m;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  unsigned long long x, n;
  vector<unsigned long long> pr;
  cin >> x >> n;
  if (x % 2 == 0) pr.push_back(2);
  while (x % 2 == 0) x /= 2;
  for (unsigned long long i = 3; i * i <= x; i += 2) {
    if (x % i == 0) pr.push_back(i);
    while (x % i == 0) x /= i;
  }
  if (x > 2) pr.push_back(x);
  unsigned long long ans = 1;
  for (auto i : pr) {
    unsigned long long y = i;
    while (i <= n) {
      unsigned long long nd = (n / i);
      ans *= pow(y, nd, 1000000007);
      ans %= 1000000007;
      if (i > n / y) break;
      i *= y;
    }
  }
  cout << ans;
}
