#include <bits/stdc++.h>
using namespace std;
long long int k, n, i, p, x, y;
long long int ami[10000000];
long long int amina[5000000];
int main() {
  while (cin >> n >> k) {
    i = 1;
    p = 0;
    x = 0;
    y = 0;
    while (i * i <= n) {
      if (n % i == 0) {
        ami[x++] = i;
        if (i != n / i) {
          amina[y++] = n / i;
        }
      }
      i++;
    }
    while (y--) {
      ami[x++] = amina[y];
    }
    i = 0;
    if (x >= k) {
      cout << ami[k - 1] << endl;
    } else {
      cout << "-1\n";
    }
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 k, n, i, p, x, y;
long long int ami[10000000];
long long int amina[5000000];
int main() {
  while (cin >> n >> k) {
    i = 1;
    p = 0;
    x = 0;
    y = 0;
    while (i * i <= n) {
      if (n % i == 0) {
        ami[x++] = i;
        if (i != n / i) {
          amina[y++] = n / i;
        }
      }
      i++;
    }
    while (y--) {
      ami[x++] = amina[y];
    }
    i = 0;
    if (x >= k) {
      cout << ami[k - 1] << endl;
    } else {
      cout << "-1\n";
    }
  }
  return 0;
}

```