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 prime[100001];
bool c[100001];
const long long MOD = 1000000007;
long long mpow(long long a, long long b) {
  if (b == 0) return 1;
  if (b == 1) {
    return a % MOD;
  } else {
    if (b % 2 == 0) {
      return (mpow((a % MOD * a % MOD) % MOD, b / 2)) % MOD;
    } else {
      return (a % MOD * mpow(a % MOD, b - 1)) % MOD;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int idx = 0;
  for (long long i = 2; i < 100001; i++) {
    if (c[i] == false) {
      for (long long j = i * i; j < 100001; j += i) {
        c[j] = true;
      }
    }
  }
  long long x, n;
  cin >> x >> n;
  long long x2 = sqrt(x);
  for (int i = 2; i <= x2; i++) {
    if (x % i == 0 && c[i] == false) {
      prime[idx++] = i;
      while (x % i == 0) {
        x /= i;
      }
    }
  }
  if (x > x2) {
    prime[idx++] = x;
  }
  long long ans = 1;
  long long n2 = n;
  for (int i = 0; i < idx; i++) {
    long long t = prime[i];
    long long last = t;
    n2 = n;
    while (t <= n2) {
      long long dif = n2 / t;
      long long temp = mpow(prime[i], dif);
      ans *= temp;
      ans %= MOD;
      n2 /= prime[i];
    }
  }
  cout << ans;
}
