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 gcd(long long x, long long y) {
  if (!y) return x;
  return gcd(y, x % y);
}
const long long MOD = 1e9 + 7;
int inf = 1e9 + 7;
long long INF = 2e18 + 9;
long long power(long long x, long long y) {
  long long res = 1ll;
  x %= MOD;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % MOD;
    y = y >> 1ll;
    x = (x * x) % MOD;
  }
  return res;
}
long long inv(long long x) {
  long long r = 1ll, t = x, k = MOD - 2ll;
  while (k) {
    if (k & 1ll) r = (long long)r * t % MOD;
    t = (long long)t * t % MOD;
    k >>= 1;
  }
  return r;
}
bool isPrime(long long x) {
  if (x == 1) return false;
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) return false;
  }
  return true;
}
const int nax = 5e5 + 13;
void prep() {}
void solve() {}
void read() {
  long long x, n;
  cin >> x >> n;
  vector<long long> prime;
  long long tmp = x;
  for (long long i = 2; i * i <= x; i++) {
    if (tmp % i == 0) {
      prime.push_back(i);
      while (tmp % i == 0) tmp /= i;
    }
  }
  if (tmp > 1) prime.push_back(tmp);
  long long res = 1;
  for (int i = 0; i < prime.size(); i++) {
    long long dzielnik = prime[i];
    while (dzielnik <= n) {
      long long ile = n / dzielnik;
      res *= power(prime[i], ile);
      res %= MOD;
      if (LLONG_MAX / prime[i] < dzielnik) {
        break;
      }
      dzielnik *= prime[i];
    }
  }
  cout << res << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  srand(time(NULL));
  int T = 1;
  prep();
  for (int i = 1; i <= T; i++) {
    read();
    solve();
  }
  return 0;
}
