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 mod = (long long)1e9 + 7;
const int maxn = (int)1e6 + 5;
const long long INF = 0x7fffffff;
const long long inf = 0x3f3f3f3f;
const double eps = 1e-8;
void f() {}
template <typename T>
void read(T &w) {
  char c, k = 1;
  while (!isdigit(c = getchar())) c == '-' ? k = -1 : k = k;
  w = c & 15;
  while (isdigit(c = getchar())) w = w * 10 + (c & 15);
  w *= k;
}
template <typename T>
void output(T x) {
  if (x < 0) putchar('-'), x = -x;
  int ss[55], sp = 0;
  do ss[++sp] = x % 10;
  while (x /= 10);
  while (sp) putchar(48 + ss[sp--]);
}
vector<int> prim;
int a[100];
;
inline long long ksm(long long x, long long n, long long mod) {
  long long res = 1;
  while (n > 0) {
    if (n & 1) res = (res * x % mod);
    x = (x * x % mod);
    n >>= 1;
  }
  return res % mod;
}
int main() {
  f();
  long long n, x;
  read(x);
  read(n);
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      prim.emplace_back(i);
      while (x % i == 0) x /= i;
    }
  }
  if (x != 1) prim.emplace_back(x);
  long long ans = 1;
  for (auto au : prim) {
    long long k = au;
    while (k <= n) {
      ans = (ans * ksm(au, n / k, mod)) % mod;
      if (n / k < au) {
        break;
      }
      k = k * au;
    }
  }
  printf("%lld\n", ans % mod);
  return 0;
}
