#include <bits/stdc++.h>
using namespace std;
long long int n, k, i, ans;
long long int a[100008], u;
int main() {
  cin >> n >> k;
  if (n < k) {
    cout << -1 << endl;
  } else {
    ans = 0;
    long long s = sqrt(n);
    for (i = 1; i <= s; i++) {
      if (n % i == 0) {
        a[++ans] = i;
        if (k == ans) {
          cout << i << endl;
          break;
        }
      }
      u = ans;
    }
    if (ans < k) {
      if (s * s == n) {
        u--;
      }
      for (i = u; i >= 1; i--) {
        ans++;
        if (ans == k) {
          cout << n / a[i] << endl;
        }
      }
    }
    if (ans < k) {
      cout << -1 << 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;
long long int n, k, i, ans;
long long int a[100008], u;
int main() {
  cin >> n >> k;
  if (n < k) {
    cout << -1 << endl;
  } else {
    ans = 0;
    long long s = sqrt(n);
    for (i = 1; i <= s; i++) {
      if (n % i == 0) {
        a[++ans] = i;
        if (k == ans) {
          cout << i << endl;
          break;
        }
      }
      u = ans;
    }
    if (ans < k) {
      if (s * s == n) {
        u--;
      }
      for (i = u; i >= 1; i--) {
        ans++;
        if (ans == k) {
          cout << n / a[i] << endl;
        }
      }
    }
    if (ans < k) {
      cout << -1 << endl;
    }
  }
}

```