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;
bool status[10000005];
vector<long long int> v, v1;
long long int a[1000005];
void siv() {
  int N = 1000005;
  int sq = sqrt(N);
  for (int i = 4; i <= N; i += 2) status[i] = 1;
  for (int i = 3; i <= sq; i += 2) {
    if (status[i] == 0) {
      for (int j = i * i; j <= N; j += i) status[j] = 1;
    }
  }
  status[1] = 1;
  for (int i = 2; i <= N; i++) {
    if (status[i] == 0) v.push_back(i);
  }
}
long long int modularExponentiation(long long int x, long long int n) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return modularExponentiation((x * x) % (1000000007), n / 2);
  else
    return (x * modularExponentiation((x * x) % (1000000007), (n - 1) / 2)) %
           (1000000007);
}
long long int loga(long long int a, long long int b) {
  long long int sum = 0;
  while (a >= b) {
    sum += a / b;
    a /= b;
  }
  return sum;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, i, j, k, l, x;
  cin >> x >> n;
  siv();
  long long int sz = v.size();
  k = x;
  i = 0;
  while (k > 1 && i < sz) {
    if (k % v[i] == 0) {
      k /= v[i];
      a[v[i]]++;
      if (a[v[i]] == 1) v1.push_back(v[i]);
    } else
      i++;
  }
  if (k > 1) v1.push_back(k);
  sz = v1.size();
  long long int sum = 1;
  for (i = 0; i < sz; i++) {
    j = v1[i];
    k = loga(n, j);
    sum =
        ((sum % (1000000007)) * (modularExponentiation(j, k) % (1000000007))) %
        (1000000007);
  }
  cout << sum << endl;
  return 0;
}
