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 int maxn = 3e5 + 5;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double Pi = acos(-1);
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
double dpow(double a, long long b) {
  double ans = 1.0;
  while (b) {
    if (b % 2) ans = ans * a;
    a = a * a;
    b /= 2;
  }
  return ans;
}
long long quick_pow(long long x, long long y) {
  long long ans = 1;
  while (y) {
    if (y & 1) {
      ans = ans * x % mod;
    }
    x = x * x % mod;
    y >>= 1;
  }
  return ans;
}
int noprime[maxn * 10], pcnt, p[maxn * 10];
void getprime(int N) {
  pcnt = 0;
  memset(noprime, 0, sizeof(noprime));
  noprime[0] = noprime[1] = 1;
  for (int i = 2; i < N; ++i) {
    if (!noprime[i]) {
      p[pcnt++] = i;
    }
    for (int j = 0; j < pcnt && i * p[j] < N; ++j) {
      noprime[i * p[j]] = 1;
      if (i % p[j] == 0) break;
    }
  }
}
long long x, n;
vector<int> vec;
void check(long long x) {
  for (int i = 0; i < pcnt; i++) {
    if (x < p[i]) break;
    if (x % p[i] == 0) vec.push_back(p[i]);
    while (x % p[i] == 0) {
      x /= p[i];
    }
  }
  if (x == 1) return;
  int flag = 1;
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      flag = 0;
      break;
    }
  }
  if (flag) vec.push_back(x);
}
long long solve(long long num) {
  long long tmp = n / num;
  long long sum = tmp;
  while (tmp) {
    sum += tmp / num;
    tmp /= num;
  }
  return quick_pow(num, sum);
}
int main() {
  scanf("%lld%lld", &x, &n);
  getprime(1e6 + 5);
  check(x);
  int sz = vec.size();
  long long ans = 1;
  for (int i = 0; i < sz; i++) {
    ans = ans * solve(vec[i]) % mod;
  }
  printf("%lld\n", ans);
}
