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 maxn = 1e5 + 7;
const int inf = 0x3f3f3f3f;
const long long mod = 1000000007;
vector<long long> pri;
void div(long long x) {
  for (long long k = 2; k * k <= x; k++) {
    if (x % k == 0) {
      pri.push_back(k);
      while (x % k == 0) x /= k;
    }
  }
  if (x > 1) pri.push_back(x);
}
long long Ksc(long long a, long long b, long long mod) {
  a %= mod, b %= mod;
  return ((a * b -
           (long long)((long long)((long double)a / mod * b + 1e-3) * mod)) %
              mod +
          mod) %
         mod;
}
long long Ksm(long long a, long long b, long long mod) {
  long long res = 1;
  while (b) {
    if (b & 1) res = Ksc(res, a, mod);
    a = Ksc(a, a, mod);
    b >>= 1;
  }
  return res;
}
int findl(long long it, long long n) {
  long long ba = 1;
  int cnt = 0;
  while ((n / ba) >= it) {
    ba *= it;
    cnt++;
  }
  return cnt;
}
int main() {
  long long x, n;
  scanf("%lld %lld", &x, &n);
  div(x);
  long long ans = 1;
  for (long long it : pri) {
    int la = findl(it, n);
    long long mul = 1;
    for (int i = 1; i <= la; i++) mul *= it;
    long long cnt, pre = 0;
    for (int i = la; i >= 1; i--) {
      cnt = n / mul;
      ans = ans * Ksm(mul, cnt - pre, mod) % mod;
      pre = cnt;
      mul /= it;
    }
  }
  printf("%lld\n", ans);
  return 0;
}
