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;
int x;
long long n, ans, res;
int su[100005], cnt, wha[100005], ned;
bool mark[100005];
void init() {
  for (int i = 2; i <= 100000; i++) {
    if (mark[i]) continue;
    mark[i] = 1;
    su[++cnt] = i;
    for (int j = i + i; j <= 100000; j += i) {
      mark[j] = 1;
    }
  }
}
long long fast(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = 1LL * res * a % 1000000007;
    a = 1LL * a * a % 1000000007;
    b >>= 1;
  }
  return res;
}
int main() {
  int i, j;
  cin >> x >> n;
  int w = x;
  ans = 1;
  init();
  for (i = 2; i * i <= x; i++) {
    if (x % i != 0) continue;
    res = 0;
    long long w = n / i;
    while (w) {
      res += w;
      w /= i;
    }
    res = fast(i, res);
    res %= 1000000007;
    ans = ans * res % 1000000007;
    while (x % i == 0) x /= i;
  }
  if (x != 1) {
    res = 0;
    long long w = n / x;
    while (w) {
      res += w;
      w /= x;
    }
    res = fast(x, res);
    res %= 1000000007;
    ans = ans * res % 1000000007;
  }
  printf("%lld\n", ans);
  return 0;
}
