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;
long long powmodm(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b % 2 == 1) {
      ans = (ans * a) % 1000000007;
    }
    b /= 2;
    a = (a * a) % 1000000007;
  }
  return ans;
}
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
vector<long long> divisors(long long x) {
  vector<long long> d;
  for (long long i = 1; i <= sqrt(x); i++) {
    if (x % i == 0) {
      d.push_back(i);
      if (i * i != x) d.push_back(x / i);
    }
  }
  sort(d.begin(), d.end());
  return d;
}
long long g(long long x, long long p) {
  long long ans = 0;
  for (long long i = 1;; i++) {
    long long f = 0, pw = pow(p, i);
    f += x / pw;
    if (f == 0) break;
    ans += f;
  }
  return ans;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long x, n;
  cin >> x >> n;
  vector<long long> pd;
  vector<long long> d = divisors(x);
  for (long long i = 0; i < d.size(); i++) {
    if (isPrime(d[i])) pd.push_back(d[i]);
  }
  vector<long long> a;
  for (long long i = 0; i < pd.size(); i++) {
    long long y = g(n, pd[i]);
    a.push_back(y);
  }
  long long ans = 1;
  for (long long i = 0; i < a.size(); i++) {
    ans = (ans * powmodm(pd[i], a[i])) % 1000000007;
  }
  cout << ans;
  return 0;
}
