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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int64_t n, i, j, k, l, sum = 0, flag = 0, t, a[200005], ans = 0;
vector<int64_t> v;
int64_t expo(int64_t base, int64_t exponent, int64_t mod = 1000000007) {
  int64_t ans = 1;
  base %= mod;
  while (exponent != 0) {
    if (exponent & 1) ans = (1LL * ans * base) % mod;
    base = (1LL * base * base) % mod;
    exponent >>= 1;
  }
  return ans % mod;
}
int64_t inverse(int64_t first) {
  return expo(first, 1000000007 - 2, 1000000007);
}
void solve() {
  int64_t n, first;
  cin >> first >> n;
  int64_t temp = first;
  for (int64_t i = 2; i * i <= first; i++) {
    if (temp % i == 0) {
      v.push_back(i);
      while (temp % i == 0) temp /= i;
    }
  }
  if (temp > 1) {
    v.push_back(temp);
  }
  int64_t ans = 1;
  for (auto i : v) {
    int64_t num = i, cur_num = i, po = 0;
    while (cur_num <= n) {
      po += (n / cur_num);
      if (1.0l * cur_num * num > n) break;
      cur_num *= num;
      po %= (1000000007 - 1);
    }
    ans = (ans * expo(num, po, 1000000007)) % 1000000007;
  }
  cout << ans << '\n';
}
signed main() {
  int64_t test = 1;
  while (test--) {
    solve();
  }
  return 0;
}
