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;
bool isprime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long int i = 5; i * i <= n; i += 6) {
    if (n % i == 0 || n % (i + 2) == 0) return false;
  }
  return true;
}
int main() {
  long long int n, k, i;
  cin >> n >> k;
  vector<pair<long long int, long long int>> div;
  div.push_back({1, n});
  for (i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      div.push_back({i, n / i});
    }
  }
  long long int len = 2 * div.size();
  long long int limit = len;
  if (div.back().first == div.back().second) limit--;
  if (k > limit)
    cout << -1 << endl;
  else {
    if (k <= len / 2) {
      cout << div[k - 1].first << endl;
    } else {
      k = k - (len / 2);
      k = (len / 2) - k;
      if (limit < len) k--;
      cout << div[k].second << endl;
    }
  }
  cerr << "time taken : " << (float)clock() / (float)CLOCKS_PER_SEC << " secs"
       << endl;
  return 0;
}