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 mod = 1e9 + 7;
const int maxn = 1e5 + 10;
int prime[maxn], ok[maxn], len;
vector<int> v;
long long x, n;
long long quick_pow(long long x, long long y) {
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x % mod;
    x = x * x % mod;
    y >>= 1;
  }
  return res;
}
int main() {
  for (int i = 2; i <= 100000; i++) {
    if (!ok[i]) prime[len++] = i;
    for (int j = 0; j < len && (long long)prime[j] * i <= 100000; j++) {
      ok[i * prime[j]] = 1;
      if (i % prime[j] == 0) break;
    }
  }
  cin >> x >> n;
  for (int i = 0; i < len && (long long)prime[i] * prime[i] <= x; i++) {
    if (x % prime[i] == 0) {
      v.push_back(prime[i]);
      while (x % prime[i] == 0) x /= prime[i];
    }
  }
  if (x > 1) v.push_back(x);
  long long ans = 1;
  for (int i = 0; i < v.size(); i++) {
    long long cnt = v[i], num = 0, m = n;
    while (m) {
      num += m / cnt;
      m /= cnt;
    }
    ans = (ans * quick_pow(v[i], num)) % mod;
  }
  cout << ans << endl;
  return 0;
}
