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(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
int mulmd(long long a, long long b) {
  long long ret = (a * b) % 1000000007;
  return (int)ret;
}
inline int power(long long x, long long y, int p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = mulmd(res, x);
    y = y >> 1;
    x = mulmd(x, x);
  }
  return (int)res;
}
int submd(long long a, long long b) {
  long long ret = a - b;
  if (ret < 0) {
    ret += 1000000007;
  }
  return (int)ret;
}
int addmd(long long a, long long b) {
  long long ret = (a + b) % 1000000007;
  return (int)ret;
}
int invPow(long long a) { return power(a, 1000000007 - 2, 1000000007); }
const int N = 1e5 + 5;
int arr[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, x;
  cin >> x >> n;
  vector<long long> facts;
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) {
      facts.push_back(i);
      while (x % i == 0) x /= i;
    }
  }
  if (x > 1) {
    facts.push_back(x);
  }
  long long ans = 1;
  for (int i = 0; i < facts.size(); i++) {
    long long st = n / facts[i];
    while (st >= 1) {
      ans = mulmd(ans, power(facts[i], st, 1000000007));
      st /= facts[i];
    }
  }
  cout << ans;
}
