#include <bits/stdc++.h>
using namespace std;
long long arr[50000];
int main() {
  long long n, k;
  while (cin >> n >> k) {
    long long cnt = 0;
    long double sq = sqrt((long double)n);
    long long res = -1;
    bool flag = false;
    for (long long i = 1; i <= sq; ++i) {
      if (n % i == 0) {
        arr[cnt++] = i;
        if (i == sq) flag = true;
      }
    }
    long long tot = 2 * cnt - flag;
    if (k > tot) {
      cout << -1 << endl;
    } else if (k > cnt) {
      cout << n / arr[tot - k] << endl;
    } else {
      cout << arr[k - 1] << endl;
    }
  }
  return 0;
}

### Prompt

Your task is to create 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 arr[50000];
int main() {
  long long n, k;
  while (cin >> n >> k) {
    long long cnt = 0;
    long double sq = sqrt((long double)n);
    long long res = -1;
    bool flag = false;
    for (long long i = 1; i <= sq; ++i) {
      if (n % i == 0) {
        arr[cnt++] = i;
        if (i == sq) flag = true;
      }
    }
    long long tot = 2 * cnt - flag;
    if (k > tot) {
      cout << -1 << endl;
    } else if (k > cnt) {
      cout << n / arr[tot - k] << endl;
    } else {
      cout << arr[k - 1] << endl;
    }
  }
  return 0;
}

```