Problem: You are given two integers n and k. Find k-th smallest divisor of n, or report that it doesn't exist.

Divisor of n is any such natural number, that n can be divided by it without remainder.

Input

The first line contains two integers n and k (1 ≤ n ≤ 1015, 1 ≤ k ≤ 109).

Output

If n has less than k divisors, output -1.

Otherwise, output the k-th smallest divisor of n.

Examples

Input

4 2


Output

2


Input

5 3


Output

-1


Input

12 5


Output

6

Note

In the first example, number 4 has three divisors: 1, 2 and 4. The second one is 2.

In the second example, number 5 has only two divisors: 1 and 5. The third divisor doesn't exist, so the answer is -1.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename T>
T gcd(T a, T b) {
  return (b ? __gcd(a, b) : a);
}
template <typename T>
T lcm(T a, T b) {
  return (a * (b / gcd(a, b)));
}
template <typename T>
T power(T e, T n, T m) {
  T x = 1 % m, p = e;
  while (n) {
    if (n & 1) x = mod(x * p, m);
    p = mod(p * p, m);
    n >>= 1;
  }
  return x;
}
vector<long long int> arr;
int main() {
  long long int n, k;
  cin >> n >> k;
  for (long long int i = 1; i * i <= n; i++) {
    if (n % i) continue;
    arr.push_back(i);
    if (i * i != n) arr.push_back(n / i);
  }
  sort((arr).begin(), (arr).end());
  k--;
  if (arr.size() <= k)
    cout << -1 << endl;
  else
    cout << arr[k] << endl;
  return 0;
}