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;
int main() {
  long long unsigned int n, k;
  cin >> n >> k;
  long long unsigned int currentCount = 0;
  double squareRoot = sqrt(n);
  long long unsigned int uppperHut = (long long unsigned int)squareRoot;
  bool perfectSquare = (squareRoot - uppperHut == 0.0 ? true : false);
  vector<long long unsigned int> factors;
  for (long long unsigned int i = 1; i <= uppperHut; i++) {
    if (n % i == 0) {
      factors.push_back(i);
      ++currentCount;
    }
    if (currentCount == k) {
      cout << i << endl;
      return 0;
    }
  }
  long long unsigned int startingLimit;
  bool valid;
  startingLimit = currentCount - 1 - (perfectSquare ? 1 : 0);
  if (k > currentCount * 2 - (perfectSquare ? 1 : 0)) {
    valid = false;
  } else {
    valid = true;
  }
  if (valid) {
    for (long long unsigned int i = startingLimit; i >= 0; i--) {
      ++currentCount;
      if (currentCount == k) {
        cout << n / factors.at(i) << endl;
        return 0;
      }
    }
  } else {
    cout << -1 << endl;
    return 0;
  }
}