#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n, k;
  cin >> n >> k;
  vector<pair<long long, long long>> v;
  vector<long long> ans;
  set<long long> s;
  for (int i = 1; i <= sqrt(n); i++) {
    if ((n % i) == 0) v.push_back({i, (n / i)});
  }
  if ((2 * v.size()) < k)
    cout << -1;
  else {
    for (auto a : v) {
      if (s.find(a.first) == s.end()) {
        ans.push_back(a.first);
        s.insert(a.first);
      }
      if (s.find(a.second) == s.end()) {
        ans.push_back(a.second);
        s.insert(a.second);
      }
    }
    sort(ans.begin(), ans.end());
    if (ans.size() >= k)
      cout << ans[k - 1];
    else
      cout << -1;
  }
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n, k;
  cin >> n >> k;
  vector<pair<long long, long long>> v;
  vector<long long> ans;
  set<long long> s;
  for (int i = 1; i <= sqrt(n); i++) {
    if ((n % i) == 0) v.push_back({i, (n / i)});
  }
  if ((2 * v.size()) < k)
    cout << -1;
  else {
    for (auto a : v) {
      if (s.find(a.first) == s.end()) {
        ans.push_back(a.first);
        s.insert(a.first);
      }
      if (s.find(a.second) == s.end()) {
        ans.push_back(a.second);
        s.insert(a.second);
      }
    }
    sort(ans.begin(), ans.end());
    if (ans.size() >= k)
      cout << ans[k - 1];
    else
      cout << -1;
  }
  return 0;
}

```