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;
const long long MOD = 1e9 + 7;
const long long MOD2 = 998244353;
const long long N = 1e3 + 5;
const long long INF = 1e18 + 5;
const long double pi = 3.14159265359;
void SRAND() {
  auto duration = std::chrono::system_clock::now().time_since_epoch();
  auto millis =
      std::chrono::duration_cast<std::chrono::milliseconds>(duration).count();
  srand(millis);
}
long long x, n, ans = 1;
long long po(long long k, long long ti) {
  if (ti == 0) return 1;
  long long tmp = po(k, ti / 2);
  if (ti & 1) return tmp * tmp % MOD * k % MOD;
  return tmp * tmp % MOD;
}
void f(long long k) {
  long long tmp = 1, tt = 0;
  while (tmp * k <= n && tmp < INF / k) tmp *= k;
  while (tmp > 1) {
    ans *= po(tmp % MOD, (n / tmp - tt) % (MOD - 1));
    ans %= MOD;
    tt += n / tmp - tt;
    tmp /= k;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> x >> n;
  for (long long i = 2; x > 1 && i * i <= x; i++) {
    if (x % i == 0) {
      f(i);
      while (x % i == 0 && x > 1) x /= i;
    }
  }
  if (x > 1) f(x);
  cout << ans << "\n";
  return 0;
}
