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;
vector<unsigned long long> primes;
void primeFactors(unsigned long long n) {
  if (n % 2 == 0) {
    primes.push_back(2);
    while (n % 2 == 0) n = n / 2;
  }
  for (unsigned long long i = 3; i <= sqrt(n); i = i + 2) {
    if (n % i == 0) {
      primes.push_back(i);
      while (n % i == 0) n = n / i;
    }
  }
  if (n > 2) primes.push_back(n);
}
unsigned long long power_count(unsigned long long n, unsigned long long p) {
  unsigned long long x = 0;
  while (n) {
    n /= p;
    x += n;
  }
  return x;
}
unsigned long long moduloMultiplication(unsigned long long a,
                                        unsigned long long b,
                                        unsigned long long mod) {
  unsigned long long res = 0;
  a %= mod;
  while (b) {
    if (b % 2 == 1) res = (res + a) % mod;
    a = (2 * a) % mod;
    a %= mod;
    b /= 2;
  }
  return res;
}
unsigned long long exponentMod(unsigned long long A, unsigned long long B,
                               unsigned long long C) {
  if (A == 0) return 0;
  if (B == 0) return 1;
  unsigned long long y;
  if (B % 2 == 0) {
    y = exponentMod(A, B / 2, C);
    y %= C;
    y = moduloMultiplication(y, y, C);
    y %= C;
  } else {
    y = A % C;
    unsigned long long temp = exponentMod(A, B - 1, C);
    y = moduloMultiplication(y, temp, C);
    y %= C;
  }
  return y % C;
}
unsigned long long power_exp(unsigned long long x, unsigned long long y,
                             unsigned long long p) {
  unsigned 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;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  unsigned long long x, n;
  cin >> x >> n;
  primes.clear();
  primeFactors(x);
  unsigned long long ans = 1;
  for (unsigned long long i = 0; i < primes.size(); i++) {
    unsigned long long p = primes[i];
    unsigned long long pow_cnt = power_count(n, p);
    unsigned long long temp = exponentMod(p, pow_cnt, 1000000007);
    temp %= 1000000007;
    ans *= temp;
    ans %= 1000000007;
  }
  cout << ans << endl;
  return 0;
}
