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 mod = 1000000007;
void genprimes(vector<long long>& primes, long long x) {
  int trgt = sqrt(x);
  for (int i = 2; i <= trgt; i++) {
    if (x % i == 0) {
      primes.push_back(i);
      while (x % i == 0) {
        x /= i;
      }
    }
  }
  if (x > 1) {
    primes.push_back(x);
  }
}
long long power(long long a, long long b) {
  if (b == 0) {
    return 1;
  }
  if (b == 1) {
    return a;
  } else {
    long long temp = power(a, b / 2);
    temp *= temp;
    temp %= mod;
    if (b % 2) {
      temp *= a;
      temp %= mod;
    }
    return temp;
  }
}
int main() {
  long long x;
  long long n;
  cin >> x >> n;
  vector<long long> factors;
  genprimes(factors, x);
  long long ans = 1;
  for (auto a : factors) {
    vector<long long> numtimes;
    unsigned long long curr = a;
    while (curr <= n && curr > 0) {
      numtimes.push_back(n / curr);
      if (n / curr < a) break;
      curr *= a;
    }
    if (numtimes.size() == 0) continue;
    for (int i = 0; i < numtimes.size() - 1; i++) {
      numtimes[i] -= numtimes[i + 1];
    }
    curr = a;
    for (int i = 0; i < numtimes.size(); i++) {
      ans *= power(curr, (i + 1) * numtimes[i]);
      ans %= mod;
    }
  }
  cout << ans << "\n";
  return 0;
}
