#include <bits/stdc++.h>
long long num[10000000];
int main() {
  long long n, k, i, t;
  scanf("%lld %lld", &n, &k);
  bool flag = 0;
  for (t = 0, i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      if (i * i != n)
        num[t] = n / i;
      else
        flag = 1;
      t++;
      if (t == k) {
        printf("%lld\n", i);
        return 0;
      }
    }
  }
  if (!flag && k <= 2 * t)
    printf("%lld\n", num[2 * t - k]);
  else if (flag && k <= 2 * t - 1)
    printf("%lld\n", num[2 * t - 1 - k]);
  else
    printf("-1\n");
  return 0;
}

### Prompt

Generate 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 num[10000000];
int main() {
  long long n, k, i, t;
  scanf("%lld %lld", &n, &k);
  bool flag = 0;
  for (t = 0, i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      if (i * i != n)
        num[t] = n / i;
      else
        flag = 1;
      t++;
      if (t == k) {
        printf("%lld\n", i);
        return 0;
      }
    }
  }
  if (!flag && k <= 2 * t)
    printf("%lld\n", num[2 * t - k]);
  else if (flag && k <= 2 * t - 1)
    printf("%lld\n", num[2 * t - 1 - k]);
  else
    printf("-1\n");
  return 0;
}

```