#include <bits/stdc++.h>
using namespace std;
vector<int> x;
int main(int argc, char **argv) {
  long long n, k;
  while (~scanf("%lld%lld", &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\n", x[k - 1]);
    else {
      int l = x.size();
      long long sf = (long long)x[l - 1] * x[l - 1] == n;
      if (k > l * 2 - sf)
        puts("-1");
      else
        printf("%lld\n", n / (long long)x[2 * l - k - sf]);
    }
  }
  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;
vector<int> x;
int main(int argc, char **argv) {
  long long n, k;
  while (~scanf("%lld%lld", &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\n", x[k - 1]);
    else {
      int l = x.size();
      long long sf = (long long)x[l - 1] * x[l - 1] == n;
      if (k > l * 2 - sf)
        puts("-1");
      else
        printf("%lld\n", n / (long long)x[2 * l - k - sf]);
    }
  }
  return 0;
}

```