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;
using ll = long long;
const ll MOD = 1000000007LL;
ll X, N;
ll p;
vector<ll> ps;
void init() { cin >> X >> N; }
void generate_ps() {
  ll t = X;
  for (ll i = 2; i * i <= t; i++) {
    if (t % i == 0) {
      while (t % i == 0) {
        t /= i;
      }
      ps.push_back(i);
    }
  }
  if (t > 1) ps.push_back(t);
}
ll mpow(ll x, ll e) {
  if (e == 0) return 1LL;
  if (e == 1) return x;
  long long t = mpow(x, e / 2);
  t = (t * t) % MOD;
  if (e % 2 == 1) t = (t * x) % MOD;
  return t;
}
void work() {
  generate_ps();
  ll s = 1LL;
  for (ll pv : ps) {
    ll e = 0, nn = N;
    while (nn >= pv) {
      e += nn / pv;
      nn /= pv;
    }
    s = (s * mpow(pv, e)) % MOD;
  }
  cout << s << endl;
}
int main() {
  init();
  work();
  return 0;
}
