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;
const long long maxm = 1e+18;
vector<long long> prime;
long long power(long long x, long long y) {
  long long res = 1;
  x = x % 1000000007;
  while (y > 0) {
    if (y & 1) res = (res * x) % 1000000007;
    y = y >> 1;
    x = (x * x) % 1000000007;
  }
  return res;
}
void primeFactors(long long n) {
  prime.clear();
  while (n % 2 == 0) {
    prime.push_back(2);
    n = n / 2;
  }
  for (int i = 3; i <= sqrt(n); i = i + 2) {
    while (n % i == 0) {
      prime.push_back(i);
      n = n / i;
    }
  }
  if (n > 2) prime.push_back(n);
}
long long getRes(long long n, long long p) {
  long long res = 1;
  long long x[101];
  long long i = 1;
  long long j = 0;
  long long z = p;
  while (true) {
    if (z > n) break;
    x[j++] = n / z;
    if (maxm / z < p) break;
    z *= p;
    i++;
  }
  for (int i = 0; i < j - 1; i++) {
    x[i] -= x[i + 1];
  }
  long long mul = 1;
  for (int i = 0; i < j; i++) {
    long long val = power(power(p, mul), x[i]);
    res = (res * (val) % 1000000007) % 1000000007;
    mul += 1;
  }
  return res;
}
int main() {
  long long x, n;
  cin >> x >> n;
  primeFactors(x);
  vector<long long>::iterator ip;
  ip = std::unique(prime.begin(), prime.end());
  prime.resize(std::distance(prime.begin(), ip));
  long long ans = 1;
  for (int i = 0; i < prime.size(); i++) {
    long long res = getRes(n, prime[i]);
    ans = (ans * res) % 1000000007;
  }
  cout << ans << "\n";
}
