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;
class TaskA {
 public:
  void solve(istream& in, ostream& out) {
    int64_t n;
    int k;
    in >> n >> k;
    vector<int64_t> divisors;
    int64_t i = 1;
    while (i * i <= n) {
      if (n % i == 0) {
        divisors.push_back(i);
      }
      ++i;
    }
    while (i * i >= n) {
      --i;
    }
    while (i > 0) {
      if (n % i == 0) {
        divisors.push_back(n / i);
      }
      --i;
    }
    if ((int)divisors.size() >= k) {
      out << divisors[k - 1] << endl;
    } else {
      out << -1 << endl;
    }
  }
};
int main() {
  std::ios::sync_with_stdio(false);
  TaskA solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}