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;
const int N = 1e5 + 5;
const long long M = (long long)1e9 + 7;
const long long inf = (long long)1e18;
const double pi = 2 * acos(0.0);
const double esp = 1e-9;
int dx[] = {0, 0, 1, -1};
int dy[] = {1, -1, 0, 0};
vector<long long> vt, tv;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL);
  int test = 1;
  for (int cs = 1; cs <= test; ++cs) {
    long long n, k;
    cin >> n >> k;
    long long sq = sqrt(n);
    for (long long i = 1ll; i <= sq; ++i) {
      if (n % i == 0ll) {
        if (i == n / i) {
          vt.push_back(i);
        } else {
          vt.push_back(i);
          tv.push_back(n / i);
        }
      }
    }
    sort(tv.begin(), tv.end());
    if (k > (int)vt.size() + (int)tv.size()) {
      cout << "-1";
    } else {
      if (k <= (int)vt.size()) {
        cout << vt[k - 1];
      } else {
        k -= (int)vt.size();
        if (k <= (int)tv.size()) {
          cout << tv[k - 1];
        } else {
          cout << "-1";
        }
      }
    }
    cout << "\n";
  }
  return 0;
}