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 double eps = 1e-6;
const int INF = 0x3f3f3f3f, N = 2000005, M = 2000005, MOD = 1e9 + 7;
int sgn(double x) { return x < -eps ? -1 : (x < eps ? 0 : 1); }
int Rand(int x) { return rand() * rand() % x + 1; }
long long qpow(long long a, long long b) {
  long long ret = 1;
  for (; b; b >>= 1, a = a * a % MOD)
    if (b & 1) ret = ret * a % MOD;
  return ret;
}
long long a[N];
int ss;
int Solve() {
  long long x, n;
  cin >> x >> n;
  for (int i = 2; i * i <= x; i++)
    if (x % i == 0) {
      a[++ss] = i;
      while (x % i == 0) x /= i;
    }
  if (x != 1) a[++ss] = x;
  long long ans = 1;
  for (int i = 1; i <= ss; i++) {
    long long k = a[i], m = n;
    while (m / k) {
      ans = ans * qpow(a[i], m / k) % MOD;
      m /= k;
    }
  }
  cout << ans << endl;
  return 0;
}
void Pre() {}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  Pre();
  while (Solve())
    ;
  return 0;
}
