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>
const int N = 1e5 + 100;
const int INF = 1e9;
const unsigned long long mod = 1e9 + 7;
using namespace std;
unsigned long long x, n, ans = 1;
vector<unsigned long long> d;
bool isprime(unsigned long long a) {
  if (a == 1) return false;
  for (unsigned long long i = (2); (i) <= (sqrt(a)); ++(i)) {
    if (a % i == 0) return false;
  }
  return true;
}
unsigned long long qpow(unsigned long long x, unsigned long long y) {
  unsigned long long ret = 1, a = x;
  for (; y; a = a * a % mod, y >>= 1)
    if (y & 1) ret = ret * a % mod;
  return ret;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> x >> n;
  for (unsigned long long i = (1); (i) <= (sqrt(x)); ++(i)) {
    if (x % i == 0) {
      if (isprime(i)) d.push_back(i);
      if (isprime(x / i) && i * i != x) d.push_back(x / i);
    }
  }
  for (unsigned long long i = 0; (i) < (d.size()); ++(i)) {
    unsigned long long tmp = d[i], cnt = 0;
    while (1) {
      ans *= qpow(d[i], n / tmp);
      ans %= mod;
      if (n / tmp < d[i]) break;
      tmp *= d[i];
    }
  }
  cout << ans % mod;
  return 0;
}
