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 int binpow(long long int a, long long int n) {
  if (n == 0) return 1;
  if (n % 2 == 1)
    return (binpow(a, n - 1)) % 1000000007 * (a % 1000000007) % 1000000007;
  else {
    long long int b = (binpow(a, n / 2) % 1000000007);
    return ((b % 1000000007) * (b % 1000000007)) % 1000000007;
  }
}
set<long long int> s;
void find(long long int n, long long int div) {
  if (n == 1) return;
  if (div * div > n) {
    s.insert(n);
    return;
  }
  if (n % div == 0) {
    s.insert(div);
    find(n / div, div);
  } else if (div == 2)
    find(n, div + 1);
  else
    find(n, div + 2);
}
int main() {
  long long int n, x;
  long long int ans = 0;
  cin >> x >> n;
  find(x, 2);
  vector<long long int> kek;
  for (auto now : s) {
    long long int r = 0;
    long long int z = now;
    while (now <= n) {
      r += n / now;
      if (n / now < z) break;
      now = now * z;
    }
    kek.push_back(r);
  }
  vector<long long int> lul;
  for (auto now : s) lul.push_back(now);
  vector<long long int> answer(s.size());
  for (int i = 0; i < s.size(); i++) {
    answer[i] = (binpow(lul[i], kek[i])) % 1000000007;
  }
  long long int an = 1;
  for (int i = 0; i < s.size(); i++)
    an = ((an % 1000000007) * (answer[i] % 1000000007)) % 1000000007;
  cout << an << endl;
  return 0;
}
