#include <bits/stdc++.h>
using namespace std;
long long n, k;
long long prime[1000005];
long long tot = 0;
void Pre(long long x) {
  long long p = sqrt(x);
  for (long long i = 1; i <= p; i++) {
    if (x % i == 0) {
      prime[++tot] = i;
      if (i != (x / i)) prime[++tot] = x / i;
    }
  }
}
signed main() {
  scanf("%lld%lld", &n, &k);
  Pre(n);
  sort(prime + 1, prime + 1 + tot);
  if (tot >= k) {
    printf("%lld\n", prime[k]);
  } else {
    printf("-1\n");
  }
  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>
using namespace std;
long long n, k;
long long prime[1000005];
long long tot = 0;
void Pre(long long x) {
  long long p = sqrt(x);
  for (long long i = 1; i <= p; i++) {
    if (x % i == 0) {
      prime[++tot] = i;
      if (i != (x / i)) prime[++tot] = x / i;
    }
  }
}
signed main() {
  scanf("%lld%lld", &n, &k);
  Pre(n);
  sort(prime + 1, prime + 1 + tot);
  if (tot >= k) {
    printf("%lld\n", prime[k]);
  } else {
    printf("-1\n");
  }
  return 0;
}

```