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 F(long long N, long long P) {
  if (P == 0) return 1;
  if (P % 2 == 0) {
    long long ret = F(N, P / 2);
    return ((ret % 1000000007) * (ret % 1000000007)) % 1000000007;
  } else
    return ((N % 1000000007) * (F(N, P - 1) % 1000000007)) % 1000000007;
}
long long prime(long long n) {
  if (n == 1) {
    return 0;
  }
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      return 0;
    }
  }
  return 1;
}
int main() {
  long long n, x, i, ans = 1, n1;
  scanf("%lld %lld", &x, &n1);
  long long p[100005], ind = 0;
  if (x % 2 == 0) {
    p[ind] = 2;
    ind++;
    if (prime(x / 2) == 1 && x != 4) {
      p[ind] = x / 2;
      ind++;
    }
  }
  for (i = 3; i * i <= x; i++) {
    if (x % i == 0 && prime(i) == 1) {
      p[ind] = i;
      ind++;
    }
    if (x % i == 0 && prime(x / i) == 1 && i * i != x) {
      p[ind] = x / i;
      ind++;
    }
  }
  if (ind == 0) {
    p[ind] = x;
    ind++;
  }
  for (i = 0; i < ind; i++) {
    long long a = p[i];
    long long num = p[i];
    long long n = n1;
    long long cnt = 0;
    while (1) {
      long long f = n / num;
      cnt += f;
      if (num <= n / a) {
        num *= a;
      } else {
        break;
      }
    }
    ans = (ans * F(a, cnt)) % 1000000007;
  }
  printf("%lld", ans);
}
