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;
void print() { cout << endl; }
void prints() {}
template <typename T, typename... A>
void print(T x, A... r) {
  cout << x << " ";
  print(r...);
}
template <typename T, typename... A>
void prints(T x, A... r) {
  cout << x << " ";
  prints(r...);
}
const long long int inf = LLONG_MAX;
const long long int mod = 1e9 + 7;
const long long int N = 1e6 + 1;
template <uint32_t mod>
class Modular {
 private:
  uint32_t n;

 public:
  Modular(int64_t _n = 0) : n((_n >= 0 ? _n : mod - (-_n) % mod) % mod) {}
  uint32_t get() const { return n; }
  bool operator==(const Modular& o) const { return n == o.n; }
  bool operator!=(const Modular& o) const { return n != o.n; }
  Modular& operator+=(const Modular& o) {
    n += o.n;
    n = (n < mod ? n : n - mod);
    return *this;
  }
  Modular& operator-=(const Modular& o) {
    n += mod - o.n;
    n = (n < mod ? n : n - mod);
    return *this;
  }
  Modular& operator*=(const Modular& o) {
    n = uint64_t(n) * o.n % mod;
    return *this;
  }
  Modular& operator/=(const Modular& o) { return (*this) *= o.inv(); }
  Modular operator+(const Modular& o) const { return Modular(*this) += o; }
  Modular operator-(const Modular& o) const { return Modular(*this) -= o; }
  Modular operator*(const Modular& o) const { return Modular(*this) *= o; }
  Modular operator/(const Modular& o) const { return Modular(*this) /= o; }
  Modular pow(uint64_t b) const {
    Modular ans(1), m = Modular(*this);
    while (b) {
      if (b & 1) {
        ans *= m;
      }
      m *= m;
      b >>= 1;
    }
    return ans;
  }
  Modular inv() const {
    int32_t a = n, b = mod, u = 0, v = 1;
    while (a) {
      int32_t t = b / a;
      b -= t * a;
      swap(a, b);
      u -= t * v;
      swap(u, v);
    }
    assert(b == 1);
    return Modular(u);
  }
};
using mint = Modular<mod>;
vector<long long int> factorise(long long int num) {
  vector<long long int> ans;
  for (long long int i = 2; i * i <= num; i++) {
    if (num % i == 0) {
      num /= i;
      ans.push_back(i);
      i--;
    }
  }
  if (num > 1) ans.push_back(num);
  return ans;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int x, n;
  cin >> x >> n;
  vector<long long int> factors = factorise(x);
  mint ans = 1;
  set<long long int> f(factors.begin(), factors.end());
  for (auto p : f) {
    long long int z = n;
    while (z > 0) {
      z /= p;
      ans *= mint(p).pow(z);
    }
  }
  print(ans.get());
  return 0;
}
