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;
long long x, n;
vector<long long> pms;
vector<long long> gs;
bool nisp[100005];
const long long md = 1000000007;
inline long long qp(long long x, long long y) {
  long long rt = 1;
  for (; y; y >>= 1, (x *= x) %= md)
    if (y & 1) (rt *= x) %= md;
  return rt;
}
vector<long long> r_p;
inline bool isp(long long x) {
  if (x < 2) return 0;
  long long tt = sqrt(x);
  for (int it = 0; it < r_p.size() and r_p[it] <= tt; ++it)
    if (x % r_p[it] == 0) return 0;
  return 1;
}
int main() {
  cin >> x >> n;
  nisp[0] = 1, nisp[1] = 1;
  for (long long i = 2; i <= 100001; ++i) {
    if (!nisp[i]) {
      r_p.push_back(i);
      for (long long j = i * 2; j <= 100001; j += i) nisp[j] = 1;
    }
  }
  long long t = sqrt(x);
  for (long long i = 1; i <= t; ++i) {
    if (x % i != 0) continue;
    long long tt = x / i;
    if (isp(i)) pms.push_back(i);
    if (isp(tt) and i != tt) pms.push_back(tt);
  }
  for (long long i : pms) {
    long long sum = 0, tn = n / i;
    for (; tn; tn /= i) sum += tn;
    gs.push_back(sum);
  }
  long long ans = 1;
  for (int it = 0; it < pms.size(); ++it) (ans *= qp(pms[it], gs[it])) %= md;
  cout << ans << '\n';
}
