#include <bits/stdc++.h>
int compare(const void *a, const void *b) {
  long long la = *(long long *)a;
  long long lb = *(long long *)b;
  return la == lb ? 0 : la < lb ? -1 : 1;
}
int main() {
  static long long aa[26880];
  long long n, a;
  int k, m;
  scanf("%lld%d", &n, &k);
  m = 0;
  for (a = 1; a <= n / a; a++)
    if (n % a == 0) {
      aa[m++] = a;
      if (a != n / a) aa[m++] = n / a;
    }
  if (k > m) {
    printf("-1\n");
    return 0;
  }
  qsort(aa, m, sizeof *aa, compare);
  printf("%lld\n", aa[k - 1]);
  return 0;
}

### Prompt

Your task is to create 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>
int compare(const void *a, const void *b) {
  long long la = *(long long *)a;
  long long lb = *(long long *)b;
  return la == lb ? 0 : la < lb ? -1 : 1;
}
int main() {
  static long long aa[26880];
  long long n, a;
  int k, m;
  scanf("%lld%d", &n, &k);
  m = 0;
  for (a = 1; a <= n / a; a++)
    if (n % a == 0) {
      aa[m++] = a;
      if (a != n / a) aa[m++] = n / a;
    }
  if (k > m) {
    printf("-1\n");
    return 0;
  }
  qsort(aa, m, sizeof *aa, compare);
  printf("%lld\n", aa[k - 1]);
  return 0;
}

```