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;
int64_t const base = 1e9 + 7;
vector<int> Primes;
map<int64_t, int64_t> mp;
map<int64_t, int64_t>::iterator it;
void Fact(int64_t first) {
  for (int64_t i = 2; i * i <= first; i++) {
    if (first % i == 0) {
      mp[i] = 0;
      while (first % i == 0) {
        first /= i;
      }
    }
  }
  if (first != 1) mp[first] = 1;
}
int64_t Pow(int64_t a, int64_t b) {
  if (b == 0) return 1LL % base;
  int64_t tmp = Pow(a, b / 2);
  tmp = (tmp * tmp) % base;
  if (b % 2 == 0)
    return tmp;
  else
    return tmp * a % base;
}
int64_t poww(int64_t a, int64_t b) {
  if (b == 0) return 1LL;
  int64_t tmp = Pow(a, b / 2);
  tmp = (tmp * tmp);
  if (b % 2 == 0)
    return tmp;
  else
    return tmp * a;
}
int main() {
  ios_base::sync_with_stdio(0);
  int64_t first;
  int64_t n;
  cin >> first >> n;
  Fact(first);
  int64_t ans = 1;
  for (it = mp.begin(); it != mp.end(); it++) {
    int64_t pre = n, cur = 0, k = (*it).first;
    int64_t s = k;
    int64_t cnt = 0, _n = n;
    cerr << "k"
         << " = " << k << '\n';
    ;
    while (_n >= k) {
      _n /= k;
      cnt++;
    }
    for (int i = 1; i <= cnt; i++) {
      cur = n / poww(k, i);
      ans = (ans * Pow(Pow(k, i - 1), pre - cur) + base) % base;
      pre = cur;
    }
    ans = (ans * (Pow(Pow(k, cnt), cur)) + base) % base;
  }
  cout << ans;
}
