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 = 1e9 + 7;
const double EPS = 1e-10;
const int INF = 1e9 + 1;
struct Mod {
  long long x;
  Mod(long long x = 0) : x(x) {}
  Mod operator+(Mod b) { return Mod((x + b.x) % MOD); }
  Mod operator-(Mod b) { return Mod((x - b.x + MOD) % MOD); }
  Mod operator*(Mod b) { return Mod((x * b.x) % MOD); }
  Mod operator/(Mod b) { return *this * inverse(b); }
  Mod operator%(Mod b) { return Mod(x % b.x); }
  static long long euclid(long long a, long long b, long long &x,
                          long long &y) {
    if (b) {
      long long d = euclid(b, a % b, y, x);
      y -= a / b * x;
      return d;
    }
    x = 1;
    y = 0;
    return a;
  }
  Mod inverse(Mod a) {
    long long x, y, g = euclid(a.x, MOD, x, y);
    assert(g == 1);
    return Mod((x + MOD) % MOD);
  }
  Mod operator^(long long e) {
    if (!e) return Mod(1);
    Mod r = *this ^ (e / 2);
    r = r * r;
    return e & 1 ? *this * r : r;
  }
};
template <typename A, typename B>
istream &operator>>(istream &in, pair<A, B> &p) {
  in >> p.first >> p.second;
  return in;
}
template <typename A, typename B>
ostream &operator<<(ostream &out, pair<A, B> &p) {
  out << p.first << " " << p.second;
  return out;
}
template <typename I>
istream &operator>>(istream &in, vector<I> &a) {
  for (auto &i : (a)) in >> i;
  return in;
}
template <typename I>
ostream &operator<<(ostream &out, vector<I> &a) {
  for (int i = 0; i < ((int)(a).size()); ++i) {
    if (i) out << " ";
    out << a[i];
  }
  return out;
}
void fast_io() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin.exceptions(cin.failbit);
}
long long mpow(long long b, long long e, long long m) {
  if (e == 0) return 1;
  long long r = mpow(b * b % m, e / 2, m);
  return e & 1 ? r * b % m : r;
}
void solve_test() {
  long long x, n;
  cin >> x >> n;
  vector<long long> primes;
  int xx = x;
  for (int i = 2; i * i <= xx; ++i) {
    if (xx % i == 0) {
      primes.push_back(i);
      while (xx % i == 0) xx /= i;
    }
  }
  if (xx > 1) {
    primes.push_back(xx);
  }
  long long ans = 1;
  for (auto &p : (primes)) {
    long long pp = p;
    while (n / pp >= p) {
      ans = ans * mpow(p, n / pp, MOD) % MOD;
      pp *= p;
    }
    ans = ans * mpow(p, n / pp, MOD) % MOD;
  }
  cout << ans << endl;
}
void setup() {}
int main() {
  setup();
  srand(time(0));
  fast_io();
  int t = 1;
  while (t--) {
    solve_test();
  }
}
