#include <bits/stdc++.h>
long long int a[10000000], b[10000000];
int main() {
  long long int k, n, i, s, j = 0, sq, x, l = 0, sum;
  scanf("%lld %lld", &n, &k);
  sq = sqrt(n);
  for (i = 1; i <= sq; i++) {
    if (n % i == 0) {
      x = n / i;
      a[j] = i;
      j++;
      if (i == x) break;
      b[l] = x;
      l++;
    }
  }
  sum = l + j;
  if (k > sum) {
    printf("-1\n");
    return 0;
  }
  if (k <= j) {
    printf("%lld\n", a[k - 1]);
  }
  if (k > j) {
    printf("%lld\n", b[sum - k]);
  }
  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>
long long int a[10000000], b[10000000];
int main() {
  long long int k, n, i, s, j = 0, sq, x, l = 0, sum;
  scanf("%lld %lld", &n, &k);
  sq = sqrt(n);
  for (i = 1; i <= sq; i++) {
    if (n % i == 0) {
      x = n / i;
      a[j] = i;
      j++;
      if (i == x) break;
      b[l] = x;
      l++;
    }
  }
  sum = l + j;
  if (k > sum) {
    printf("-1\n");
    return 0;
  }
  if (k <= j) {
    printf("%lld\n", a[k - 1]);
  }
  if (k > j) {
    printf("%lld\n", b[sum - k]);
  }
  return 0;
}

```