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;
long long m = 1000000007;
long long M = 1000000000000000000;
long long Logn(long long n, long long r) {
  return (n > r - 1) ? 1 + Logn(n / r, r) : 0;
}
long long power1(long long p, long long b) {
  long long r = 1;
  while (p) {
    if (p & 1) r = r * b;
    b = b * b;
    p = p / 2;
  }
  return r;
}
long long power2(long long p, long long b) {
  long long r = 1;
  while (p) {
    if (p & 1) r = (r % m) * (b % m) % m;
    b = (b % m) * (b % m);
    p = p / 2;
  }
  return r % m;
}
int main(void) {
  long long x, n;
  cin >> x >> n;
  vector<long long> prime;
  for (long long i = 2; i <= sqrt(x); i++) {
    if (x % i == 0) {
      prime.push_back(i);
      while (x % i == 0) x = x / i;
    }
  }
  if (x > 1) prime.push_back(x);
  long long ans = 1, count;
  for (long long i = 0; i < prime.size(); i++) {
    long long m1 = Logn(n, prime[i]);
    long long num1 = power1(m1, prime[i]);
    while (n / num1 == prime[i]) num1 = num1 * prime[i];
    count = 0;
    while (m1--) {
      long long num2 = n / num1;
      num2 = num2 - count;
      long long num3 = power2(num2, num1) % m;
      ans = (ans % m) * (num3 % m) % m;
      count = count + num2;
      num1 = num1 / prime[i];
    }
  }
  ans = ans % m;
  cout << ans << endl;
  return 0;
}
