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>
const int fx[] = {+1, -1, +0, +0};
const int fy[] = {+0, +0, +1, -1};
using namespace std;
bool sign[100010 + 2];
vector<long long> prime;
void Prime() {
  prime.push_back(2ll);
  memset(sign, true, sizeof sign);
  for (long long i = 3ll; i * i <= 100010; i += 2ll) {
    if (sign[i])
      for (long long j = i * i; j <= 100010; j += (i * 2ll)) {
        sign[j] = false;
      }
  }
  for (long long i = 3ll; i <= 100010; i += 2ll)
    if (sign[i]) prime.push_back(i);
}
vector<long long> v;
void Factor(long long num) {
  for (long long i = 0ll; prime[i] * prime[i] <= num && i < prime.size(); i++) {
    if (num % prime[i] == 0ll) {
      v.push_back(prime[i]);
      while (num % prime[i] == 0LL) num /= prime[i];
    }
  }
  if (num > 1) v.push_back(num);
}
long long Big(long long b, long long p) {
  if (p == 0) return 1;
  long long x = Big(b, p / 2);
  x = (x * x) % 1000000007;
  if (p % 2LL == 1LL) x = (x * b) % 1000000007;
  return x % 1000000007;
}
long long TT(long long temp, long long n) {
  long long tt = temp;
  long long ans = 0;
  while (n >= tt) {
    ans += (n / tt);
    if (n / tt < temp) break;
    tt = (tt * temp);
  }
  return ans;
}
int main() {
  long long x, n;
  cin >> x >> n;
  Prime();
  Factor(x);
  long long ans = 1ll;
  for (int i = 0; i < v.size(); i++) {
    long long t2 = TT(v[i], n);
    ans = (ans * (Big(v[i], t2))) % 1000000007;
  }
  cout << ans << endl;
  return 0;
}
