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 int MX = 1e5 + 50;
const long long MOD = 1e9 + 7;
long long n, x;
int len = 0;
long long prime[MX];
bool is_prime[MX];
long long qadd(long long x, long long p) {
  long long ans = 0;
  while (p) {
    if (p & 1) ans = (x + ans) % MOD;
    x = (x + x) % MOD;
    p >>= 1;
  }
  return ans;
}
long long qpow(long long x, long long p) {
  long long ans = 1;
  while (p) {
    if (p & 1) ans = qadd(ans, x);
    x = qadd(x, x);
    p >>= 1;
  }
  return ans;
}
void init() {
  memset(is_prime, true, sizeof(is_prime));
  for (int i = 2; i < MX; i++) {
    if (is_prime[i]) {
      prime[len++] = i;
      for (int j = i * 2; j < MX; j += i) {
        is_prime[j] = false;
      }
    }
  }
}
long long p[MX];
int plen = 0;
void make_p() {
  for (int i = 0; i < len && x != 0; i++) {
    if (x % prime[i] == 0)
      p[plen++] = prime[i];
    else
      continue;
    while (x % prime[i] == 0) {
      x /= prime[i];
    }
  }
  if (x != 1) p[plen++] = x;
}
int main() {
  ios::sync_with_stdio(true);
  init();
  cin >> x >> n;
  make_p();
  long long ans = 1;
  for (int i = 0; i < plen; i++) {
    long long t = p[i], cnt = 0, pre = 0;
    while (t <= n / p[i]) t *= p[i];
    while (t / p[i] != 0) {
      long long mi = n / t - pre;
      ans = ans * qpow(t, mi) % MOD;
      pre = n / t;
      t /= p[i];
    }
  }
  cout << ans << endl;
}
