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;
vector<long long> primes;
int prime[32000];
long long Power(long long a, long long b) {
  if (b == 0) return 1;
  long long temp = Power(a, b / 2);
  long long ans = (temp * temp) % 1000000007;
  if (b % 2 == 0) return ans;
  ans *= (a % 1000000007);
  ans %= 1000000007;
  return ans;
}
vector<long long> primeFactorOf(long long x) {
  vector<long long> ans;
  for (long long i = 0; i < primes.size() && primes[i] <= sqrt(x); i++) {
    long long temp = primes[i];
    if (x % temp == 0) {
      ans.push_back(temp);
      while (x % temp == 0) {
        x /= temp;
      }
    }
  }
  if (x > 1) ans.push_back(x);
  return ans;
}
int main() {
  prime[1] = 1;
  prime[0] = 1;
  for (long long i = 2; i <= 31625; i++) {
    if (prime[i] == 0) {
      primes.push_back(i);
      for (long long j = i * 2; j <= 31625; j += i) {
        prime[j] = 1;
      }
    }
  }
  long long m, n;
  cin >> m >> n;
  unsigned long long ans = 1;
  vector<long long> p = primeFactorOf(m);
  for (unsigned long long i : p) {
    unsigned long long j = 1;
    double x = (log2(n) / log2(i)) + 0.0000000000001;
    long long t = x;
    for (long long k = 1; k <= t; k++) {
      long long temp = 0;
      j *= i;
      if ((unsigned long long)n / j >= i) {
        temp += ((unsigned long long)n / (unsigned long long)(j * i));
      }
      long long cnt = n / j - temp;
      ans *= Power(j, cnt);
      ans %= 1000000007;
    }
  }
  cout << ans << endl;
}
