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 NN = 100100;
const long long inf = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-6;
int sgn(double x) {
  if (fabs(x) < eps) return 0;
  if (x > 0) return 1;
  return -1;
}
const long long G = 1e9 + 7;
long long n, m, k, x, y, z, cnt, tmp, sum, num, siz;
int a[NN], vis[NN], c[NN * 10], b[NN];
char s[5000];
const double e = 2.718281828459;
const double pi = acos(-1.0);
long long poww(long long x, long long y) {
  if (y == 0) return 1;
  long long tmp = poww(x, y / 2);
  tmp = tmp * tmp % G;
  if (y & 1) tmp *= x;
  return tmp % G;
}
long long solve(long long x) {
  long long tmp = n / x, cnt = 0;
  while (tmp) {
    cnt += tmp;
    tmp /= x;
  }
  return poww(x, cnt);
}
long long pri[NN];
void prime() {
  for (int i = 2; i <= 1000000; i++) {
    if (c[i] == 0) b[++siz] = i;
    for (int j = 1; j <= siz && b[j] * i <= 1000000; j++) {
      c[b[j] * i] = 1;
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  prime();
  cin >> x >> n;
  for (int i = 1; i <= siz; i++) {
    if (x % b[i] == 0) pri[++num] = b[i];
    while (x % b[i] == 0) x /= b[i];
  }
  if (x > 1) pri[++num] = x;
  long long ans = 1;
  for (int i = 1; i <= num; i++) {
    ans *= solve(pri[i]);
    ans %= G;
  }
  cout << ans << endl;
}
