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;
int d4x[4] = {1, 0, -1, 0};
int d4y[4] = {0, 1, 0, -1};
int d8x[8] = {0, 1, 1, 1, 0, -1, -1, -1};
int d8y[8] = {1, 1, 0, -1, -1, -1, 0, 1};
const long long oo = 1e18;
const long long maxN = 1e6;
const long long MOD = 1e9 + 7;
void maximize(int &a, int b) { a = max(a, b); }
void minimize(int &a, int b) { a = min(a, b); }
long long Power(long long a, long long b) {
  if (b == 0) return 1;
  if (b == 1) return a % MOD;
  long long tmp = Power(a, b / 2) % MOD;
  if (b % 2) return tmp * tmp % MOD * a % MOD;
  return tmp * tmp % MOD;
}
void solve() {
  long long x, n, ans = 1;
  vector<long long> prime;
  cin >> x >> n;
  long long t = 2;
  while (sqrt(x) >= t) {
    if (x % t == 0) {
      prime.push_back(t);
      while (x % t == 0) {
        x /= t;
      }
    }
    t++;
  }
  if (x > 1) prime.push_back(x);
  for (int i = 0; i < prime.size(); i++) {
    long long now = prime[i];
    while (now <= n) {
      long long tmp = n / now;
      tmp = tmp - (tmp / prime[i]);
      ans *= Power(now % MOD, tmp) % MOD;
      ans %= MOD;
      if (now <= (n / prime[i]) + 1) {
        now *= prime[i];
      } else
        break;
    }
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
}
