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>
const int N = 1e7;
const int mod = 1e9 + 7;
using namespace std;
void Ferrari() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
bool isprime(int x) {
  if (x == 0 || x == 1) return 0;
  for (int i = 2; i <= x / i; ++i)
    if (x % i == 0) return 0;
  return 1;
}
long long fastpow(long long a, long long b) {
  a %= mod;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
    b >>= 1;
  }
  return res;
}
int main() {
  Ferrari();
  long long ans = 1, x, n;
  cin >> x >> n;
  vector<int> p;
  for (int i = 1; i <= x / i; ++i) {
    if (x % i == 0) {
      if (isprime(i)) p.push_back(i);
      if (i * i != x && isprime(x / i)) p.push_back(x / i);
    }
  }
  for (int i = 0; i < (int)(p.size()); ++i) {
    long long pre = 0;
    long long val = 1;
    for (; (val * p[i] / p[i]) == val && val * p[i] <= n;) {
      val *= p[i];
    }
    for (; val >= p[i];) {
      long long cnt = n / val - pre;
      ans = ans * fastpow(val, cnt);
      ans %= mod;
      pre += cnt;
      val /= p[i];
    }
  }
  cout << ans % mod;
  return 0;
}
