#include <bits/stdc++.h>
using namespace std;
vector<int> x;
int main() {
  long long n;
  int k;
  scanf("%lld%d", &n, &k);
  for (long long i = 1; i * i <= n; ++i)
    if (n % i == 0) x.push_back(i);
  if (x.size() >= k)
    printf("%d", x[k - 1]);
  else {
    int l = x.size();
    int rslt = (long long)x[l - 1] * x[l - 1] == n;
    if (k > l * 2 - rslt)
      printf("-1");
    else
      printf("%lld", n / (long long)x[2 * l - k - rslt]);
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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;
vector<int> x;
int main() {
  long long n;
  int k;
  scanf("%lld%d", &n, &k);
  for (long long i = 1; i * i <= n; ++i)
    if (n % i == 0) x.push_back(i);
  if (x.size() >= k)
    printf("%d", x[k - 1]);
  else {
    int l = x.size();
    int rslt = (long long)x[l - 1] * x[l - 1] == n;
    if (k > l * 2 - rslt)
      printf("-1");
    else
      printf("%lld", n / (long long)x[2 * l - k - rslt]);
  }
  return 0;
}

```