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 M = 1e9 + 7;
const long long MAXN = 1e6 + 7;
const long double pi = acos(-1);
inline long long read() {
  long long x = 0, ff = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') ff = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return x * ff;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
std::vector<long long> prime;
bool is_composite[MAXN];
void sieve(long long n) {
  std::fill(is_composite, is_composite + n, false);
  for (long long i = 2; i < n; ++i) {
    if (!is_composite[i]) prime.push_back(i);
    for (long long j = 0; j < prime.size() && i * prime[j] < n; ++j) {
      is_composite[i * prime[j]] = true;
      if (i % prime[j] == 0) break;
    }
  }
}
long long pw(long long x, long long y) {
  if (!y) return 1;
  x %= M;
  long long ans = pw(x, y / 2);
  ans = (ans * ans) % M;
  if (y + 1 & 1) return ans;
  return (ans * x) % M;
}
set<long long> fac(long long n) {
  set<long long> s;
  for (; n % 2 < 1; n /= 2) s.insert(2);
  for (long long i = 3; i * i <= n; i += 2)
    for (; n % i < 1; n /= i) s.insert(i);
  if (n > 2) s.insert(n);
  return s;
}
void doit() {
  long long x = read();
  long long n = read();
  set<long long> s = fac(x);
  vector<long long> v;
  long long ans = 1;
  for (auto i : s) {
    long long a = 0;
    for (long long p = i;; p *= i) {
      a += n / p;
      a %= (M - 1);
      if (p > (long double)n / i) break;
    }
    ans *= pw(i, a);
    ans %= M;
  }
  cout << ans;
}
int main() {
  doit();
  return 0;
}
