#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n;
  cin >> n;
  long k;
  cin >> k;
  long long i;
  long long arr[100000];
  long long count = 0, inc = 0, n2 = n, n1 = sqrt(n);
  for (i = 1; i <= n1; i++) {
    if (n2 % i == 0) {
      arr[++inc] = i;
    }
  }
  int t;
  if (n1 * n1 == n)
    t = inc - 1;
  else
    t = inc;
  for (i = t; i >= 1; i--) {
    arr[++inc] = n / arr[i];
  }
  if (k > inc)
    cout << "-1";
  else
    cout << arr[k];
  return 0;
}

### 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;
  cin >> n;
  long k;
  cin >> k;
  long long i;
  long long arr[100000];
  long long count = 0, inc = 0, n2 = n, n1 = sqrt(n);
  for (i = 1; i <= n1; i++) {
    if (n2 % i == 0) {
      arr[++inc] = i;
    }
  }
  int t;
  if (n1 * n1 == n)
    t = inc - 1;
  else
    t = inc;
  for (i = t; i >= 1; i--) {
    arr[++inc] = n / arr[i];
  }
  if (k > inc)
    cout << "-1";
  else
    cout << arr[k];
  return 0;
}

```