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 long long int MOD = 1e9 + 7;
const long long int INF = 1000111000111000111LL;
const long long int MAX = 2e5 + 5;
const long double PI = 3.1415926536;
long long int modular_exp(long long int base, long long int exp,
                          long long int mod) {
  long long int ans = 1;
  while (exp) {
    if (exp % 2) ans = ans % mod * base % mod;
    base = base % mod * base % mod;
    base %= mod;
    exp /= 2;
    ans %= mod;
  }
  return ans;
}
long long int inverse_modulo(long long int base, long long int mod) {
  return modular_exp(base, mod - 2, mod);
}
int32_t main() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int x, n;
  cin >> x >> n;
  set<long long int> prime;
  long long int ans = 1;
  for (long long int i = 2; i <= sqrt(x); i++) {
    if (x % i == 0) {
      while (x % i == 0 and x) {
        x /= i;
      }
      prime.insert(i);
    }
    if (x == 0) {
      break;
    }
  }
  if (x != 1) {
    prime.insert(x);
  }
  for (auto it : prime) {
    long long int tmp = n;
    long long int cnt = 0;
    while (tmp) {
      cnt += (tmp / it);
      tmp /= it;
      cnt %= (MOD - 1);
    }
    ans = (ans % MOD) * ((modular_exp(it, cnt, MOD)) % MOD);
    ans %= MOD;
  }
  cout << (ans + MOD) % MOD;
  return 0;
}
