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 p[1000000], prime;
void desc(long long x) {
  long long d = 2;
  while (d * d <= x) {
    if (x % d) {
      d++;
      continue;
    }
    p[++prime] = d;
    while (x % d == 0) {
      x /= d;
    }
    d++;
  }
  if (x > 1) {
    p[++prime] = x;
  }
}
long long calc(long long n, long long y) {
  long long put = y;
  long long sum = 0;
  while (n >= put) {
    sum = (sum + n / put);
    if (n / put < y) return sum;
    put *= y;
  }
  return sum;
}
long long lgput(long long a, long long b) {
  long long p(1);
  long long aa(a);
  while (b) {
    if (b & 1) {
      p = p * aa;
      p %= 1000000007;
    }
    aa = (aa * aa) % 1000000007;
    b >>= 1;
  }
  return p;
}
long long sol = 1;
int32_t main() {
  long long n, x, nw, sux;
  scanf("%lld %lld", &x, &n);
  desc(x);
  for (long long i = 1; i <= prime; ++i) {
    nw = (calc(n, p[i]));
    sux = lgput(p[i], nw);
    sol = ((sol % 1000000007) * sux % 1000000007) % 1000000007;
  }
  printf("%lld", sol % 1000000007);
  return 0;
}
