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 double EPS = 1e-10;
using ll = long long;
using Pi = pair<int, int>;
using Pl = pair<ll, ll>;
vector<pair<ll, int>> factorize(ll n) {
  vector<pair<ll, int>> div;
  ll d = n;
  for (ll i = 2; i * i <= n; i++) {
    pair<ll, int> p;
    if (d % i == 0) {
      p = make_pair(i, 0);
      while (d % i == 0) {
        d /= i;
        p.second++;
      }
      div.emplace_back(p);
    }
  }
  if (d != 1) div.emplace_back(d, 1);
  return div;
}
ll modpow(ll a, ll n, ll m) {
  if (n == 0) return 1;
  ll half = modpow(a, n / 2, m);
  ll res = half * half % m;
  if (n & 1) res = res * a % m;
  return res;
}
ll calc(ll n, ll k) {
  ll res = 0;
  while (n % k == 0) {
    n /= k;
    res++;
  }
  return res;
}
int main() {
  ll x, n;
  cin >> x >> n;
  vector<pair<ll, int>> div = factorize(x);
  ll ans = 1;
  for (auto u : div) {
    ll k = u.first;
    ll cnt = n / k;
    ll d = k;
    while (d <= n && d <= (double)INT64_MAX / k) {
      d *= k;
      cnt += n / d;
    }
    ans *= modpow(k, cnt, 1000000007);
    ans %= 1000000007;
  }
  cout << ans << endl;
}
