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;
using namespace std;
long long power(long long a, long long b, long long c) {
  long long res = 1;
  while (b > 0) {
    if (b % 2 == 1) res = (res * a) % c;
    a = (a * a) % c;
    b = b / 2;
  }
  return res;
}
void pf(long long n, unordered_set<long long>& s) {
  while (n % 2 == 0) {
    s.insert(2);
    n = n / 2;
  }
  for (long long i = 3; i <= sqrt(n); i = i + 2) {
    while (n % i == 0) {
      s.insert(i);
      n = n / i;
    }
  }
  if (n > 2) s.insert(n);
}
long long dig(long long a) {
  long long c = 0;
  while (a != 0) {
    c++;
    a = a / 10;
  }
  return c;
}
long long pxf(long long n, long long a) {
  long long ans = 0, p = a;
  while (1) {
    ans = ans + n / p;
    if (dig(p) + dig(a) >= 20) return ans;
    p = p * a;
  }
  return ans;
}
void slove() {
  long long ans = 1, x, n;
  cin >> x >> n;
  unordered_set<long long> s;
  pf(x, s);
  for (auto a : s) {
    ans = (ans * power(a, pxf(n, a), 1000000007)) % 1000000007;
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  slove();
}
