#include <bits/stdc++.h>
using namespace std;
long long n;
long long k;
int main(void) {
  while (~scanf("%lld %lld", &n, &k)) {
    long long t = (long long)sqrt(n);
    long long i;
    long long cnt = 0;
    for (i = 1; i <= t; ++i) {
      if (n % i == 0) ++cnt;
      if (cnt == k) {
        printf("%lld\n", i);
        break;
      }
    }
    if (cnt != k) {
      --i;
      if (t * t == n) --i;
      for (; i >= 1; --i) {
        if (n % i == 0) ++cnt;
        if (cnt == k) {
          printf("%lld\n", n / i);
          break;
        }
      }
      if (cnt != k) printf("-1\n");
    }
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 n;
long long k;
int main(void) {
  while (~scanf("%lld %lld", &n, &k)) {
    long long t = (long long)sqrt(n);
    long long i;
    long long cnt = 0;
    for (i = 1; i <= t; ++i) {
      if (n % i == 0) ++cnt;
      if (cnt == k) {
        printf("%lld\n", i);
        break;
      }
    }
    if (cnt != k) {
      --i;
      if (t * t == n) --i;
      for (; i >= 1; --i) {
        if (n % i == 0) ++cnt;
        if (cnt == k) {
          printf("%lld\n", n / i);
          break;
        }
      }
      if (cnt != k) printf("-1\n");
    }
  }
  return 0;
}

```