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 + 10;
const long long MOD = 1e9 + 7;
long long inline in() {
  long long x = 0, flag = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') flag = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + (ch ^ 48), ch = getchar();
  return x * flag;
}
long long n, x;
long long tot, p[MAXN];
bool flag[MAXN];
void filter() {
  for (int i = 2; i < MAXN; ++i) {
    if (!flag[i]) p[++tot] = i;
    for (int j = 1; j <= tot && i * p[j] < MAXN; ++j) {
      flag[i * p[j]] = true;
      if (i % p[j] == 0) break;
    }
  }
}
int cnt;
long long rec[MAXN];
long long ans = 1;
long long fpow(long long a, long long b) {
  long long ret = 1;
  while (b) {
    if (b % 2 == 1)
      b--, (ret *= a) %= MOD;
    else
      b /= 2, (a *= a) %= MOD;
  }
  return ret;
}
int main() {
  x = in();
  n = in();
  filter();
  for (int i = 1; i <= tot; ++i) {
    if (x % p[i] == 0) rec[++cnt] = p[i];
    while (x % p[i] == 0 && x != 0) x /= p[i];
  }
  if (x != 1) rec[++cnt] = x;
  for (int i = 1; i <= cnt; ++i) {
    for (long long now = rec[i]; now <= n; now *= rec[i]) {
      (ans *= fpow(now % MOD, n / now - n / now / rec[i])) %= MOD;
      if (n / now < rec[i]) break;
    }
  }
  printf("%lld\n", ans);
  return 0;
}
