#include <bits/stdc++.h>
using namespace std;
long long n, k, cnt, s[10000001];
signed main() {
  scanf("%lld%lld", &n, &k);
  if (n == 1) {
    if (k == 1)
      printf("1");
    else
      printf("-1");
    return 0;
  }
  for (long long i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      s[++cnt] = i;
      if (i != n / i) s[++cnt] = n / i;
    }
  }
  sort(s + 1, s + cnt + 1);
  if (cnt < k)
    printf("-1");
  else
    printf("%lld", s[k]);
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
long long n, k, cnt, s[10000001];
signed main() {
  scanf("%lld%lld", &n, &k);
  if (n == 1) {
    if (k == 1)
      printf("1");
    else
      printf("-1");
    return 0;
  }
  for (long long i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      s[++cnt] = i;
      if (i != n / i) s[++cnt] = n / i;
    }
  }
  sort(s + 1, s + cnt + 1);
  if (cnt < k)
    printf("-1");
  else
    printf("%lld", s[k]);
  return 0;
}

```