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;
void seive(long long n, vector<long long> &prime) {
  prime[1] = 1;
  long long i, j;
  for (i = 2; i < n; i++) {
    if (prime[i] == 1) {
      for (j = i * i; j < n; j += i) prime[j] = 0;
    }
  }
}
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;
}
bool sec(pair<long long, long long> a, pair<long long, long long> b) {
  return a.second < b.second;
}
long long pom(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long modInv(long long n, long long p) { return pom(n, p - 2, p); }
long long root(long long x, vector<long long> &id) {
  while (id[x] != x) {
    id[x] = id[id[x]];
    x = id[x];
  }
  return x;
}
void uni(long long x, long long y, vector<long long> &id) {
  long long a = root(x, id);
  long long b = root(y, id);
  id[a] = id[b];
}
long long ncr(long long n, long long r, long long p, long long fact[],
              long long mod) {
  long long x =
      fact[n] * modInv(fact[r], p) % mod * modInv(fact[n - r], p) % mod;
  cout << x;
  return x;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  srand(time(NULL));
  ;
  long long x, n, N = 1e5 + 1, i, mod = 1e9 + 7, an = 1, y, m;
  cin >> x >> n;
  m = n;
  for (i = 1; i * i <= x; i++) {
    if (x % i == 0) {
      if (isPrime(i)) {
        y = i, n = m;
        while (y <= n) {
          an = (an * pom(i, n / y, mod)) % mod;
          n /= i;
        }
      }
      if (isPrime(x / i) && x != i * i) {
        y = x / i, n = m;
        while (y <= n) {
          an = (an * pom(x / i, n / y, mod)) % mod;
          n /= (x / i);
        }
      }
    }
  }
  cout << an << endl;
}
