#include <bits/stdc++.h>
using namespace std;
int main() {
  long long i, t, k;
  scanf("%lld %lld", &t, &k);
  set<long long> ss;
  set<long long>::iterator it;
  for (i = 1; i * i <= t; i++) {
    if (t % i == 0) {
      ss.insert(i);
      ss.insert(t / i);
    }
  }
  if (k > ss.size()) {
    printf("-1\n");
  } else {
    i = 1;
    for (it = ss.begin(); it != ss.end(); it++) {
      if (i == k) {
        printf("%lld\n", *it);
        break;
      }
      i++;
    }
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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 i, t, k;
  scanf("%lld %lld", &t, &k);
  set<long long> ss;
  set<long long>::iterator it;
  for (i = 1; i * i <= t; i++) {
    if (t % i == 0) {
      ss.insert(i);
      ss.insert(t / i);
    }
  }
  if (k > ss.size()) {
    printf("-1\n");
  } else {
    i = 1;
    for (it = ss.begin(); it != ss.end(); it++) {
      if (i == k) {
        printf("%lld\n", *it);
        break;
      }
      i++;
    }
  }
  return 0;
}

```