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>
#pragma warning(disable : 4996)
using namespace std;
clock_t start_time, end_time;
void open() {
  ((void)0);
  ((void)0);
  ((void)0);
  ((void)0);
}
void close() {
  ((void)0);
  ((void)0);
}
const int MAX = 1e5 + 1;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const long long LL_INF = 0x3f3f3f3f3f3f3f3fLL;
const double PI = acos(-1);
const long double ERR = 1e-10;
const int move_r[] = {-1, 0, 1, 0, -1, -1, 1, 1};
const int move_c[] = {0, -1, 0, 1, -1, 1, -1, 1};
long long _pow(long long base, long long exp) {
  long long ret = 1;
  while (exp) {
    if (exp % 2LL) ret = ret * base % MOD;
    base = base * base % MOD;
    exp /= 2LL;
  }
  return ret;
}
long long count(long long val, long long p) {
  long long cnt = 0;
  while (val) {
    cnt += val / p;
    val /= p;
  }
  return cnt;
}
int main(int argc, char *argv[]) {
  open();
  long long x, n;
  scanf("%lld %lld", &x, &n);
  long long ans = 1;
  for (long long i = 2; i * i <= x; ++i) {
    if (x % i == 0) {
      ans = ans * _pow(i, count(n, i)) % MOD;
      while (x % i == 0) x /= i;
    }
  }
  if (x > 1) ans = ans * _pow(x, count(n, x)) % MOD;
  printf("%lld\n", ans);
  close();
  return 0;
}
