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>
#pragma GCC optimize("Ofast")
using namespace std;
const long long inf = 2e18;
const long double pi = 3.141592653589793238;
long long MOD = 1000000007;
const long long N = 2e5 + 10;
long long x, n;
long long MOD1 = MOD - 1;
long long power(long long a, long long b, long long m) {
  long long ans = 1;
  a %= m;
  a += m;
  a %= m;
  while (b) {
    if (b & 1) ans = (ans * a) % m;
    b /= 2;
    a = (a * a) % m;
  }
  return ans;
}
void solve() {
  cin >> x >> n;
  set<long long> fact;
  long long temp = x;
  for (long long i = 2; i * i <= x; i++) {
    if (x % i) continue;
    fact.insert(i);
    while (x % i == 0) x /= i;
  }
  if (x > 1) fact.insert(x);
  x = temp;
  long long ans = 1;
  for (auto i : fact) {
    long long b = 1;
    while (b <= n / i) {
      b *= i;
    }
    long long cur = 0;
    while (b > 1) {
      long long times = (n) / b;
      long long temp = times - cur;
      temp %= MOD1;
      temp += MOD1;
      temp %= MOD1;
      ans *= power(b, temp, MOD);
      ans %= MOD;
      ans += MOD;
      ans %= MOD;
      cur += temp;
      cur %= (MOD1);
      cur += (MOD1);
      cur %= (MOD1);
      b /= i;
    }
  }
  ans %= MOD;
  ans += MOD;
  ans %= MOD;
  cout << ans << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
}
