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 int N = 1e6 + 5;
const long long MOD = 1e9 + 7;
int x;
long long n;
vector<int> v;
int top, prime[N], f[N];
void init() {
  top = 0;
  for (int i = 2; i < N; i++) {
    if (!f[i]) prime[top++] = i;
    for (int j = 0; j < top && prime[j] * i < N; j++) {
      f[i * prime[j]] = 1;
      if (i % prime[j] == 0) break;
    }
  }
}
void get(int x) {
  int idx = 0;
  while (x) {
    if (idx >= top) break;
    if (x % prime[idx] == 0) {
      int a = prime[idx];
      v.push_back(a);
      while (x % a == 0) x /= a;
    }
    idx++;
  }
  if (x > 1) v.push_back(x);
  sort(v.begin(), v.end());
}
long long quick_mod(long long a, long long b, long long p) {
  long long ans = 1;
  a %= p;
  while (b) {
    if (b & 1) ans = ans * a % p;
    b >>= 1;
    a = a * a % p;
  }
  return ans;
}
void solve() {
  long long ans = 1;
  for (int i = 0; i < v.size(); i++) {
    long long p = v[i];
    while (p <= n) {
      long long q = quick_mod(v[i], n / p, MOD);
      ans = ans * q % MOD;
      if (p > n / v[i]) break;
      p = p * v[i];
    }
  }
  printf("%lld\n", ans);
}
int main() {
  init();
  scanf("%d%lld", &x, &n);
  get(x);
  solve();
  return 0;
}
