#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n, k;
  cin >> n >> k;
  vector<long long> v;
  vector<long long> v1;
  set<long long> s;
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      v.push_back(i);
      s.insert(i);
      if (s.find(n / i) == s.end()) v1.push_back(n / i);
    }
  }
  long long cnt = 0;
  long long ans = -1;
  long long i = 0;
  while (cnt < v.size()) {
    cnt++;
    if (k == cnt) {
      ans = v[i];
      break;
    }
    i++;
  }
  long long j = 0, p = 0;
  sort(v1.begin(), v1.end());
  while (cnt != k && p < v1.size()) {
    cnt++;
    if (cnt == k) {
      ans = v1[j];
      break;
    }
    j++;
    p++;
  }
  cout << ans << endl;
}

### Prompt

Please formulate 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<long long> v;
  vector<long long> v1;
  set<long long> s;
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      v.push_back(i);
      s.insert(i);
      if (s.find(n / i) == s.end()) v1.push_back(n / i);
    }
  }
  long long cnt = 0;
  long long ans = -1;
  long long i = 0;
  while (cnt < v.size()) {
    cnt++;
    if (k == cnt) {
      ans = v[i];
      break;
    }
    i++;
  }
  long long j = 0, p = 0;
  sort(v1.begin(), v1.end());
  while (cnt != k && p < v1.size()) {
    cnt++;
    if (cnt == k) {
      ans = v1[j];
      break;
    }
    j++;
    p++;
  }
  cout << ans << endl;
}

```