#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 1e6 + 50;
const int mod = 1e9 + 7;
set<long long> s;
void divide(long long num) {
  for (register long long i = 1; i * i <= num; i++) {
    if (num % i == 0) s.insert(i), s.insert(num / i);
  }
}
int main() {
  long long n, k;
  scanf("%lld%lld", &n, &k);
  divide(n);
  if (k > s.size())
    printf("-1");
  else {
    auto p = s.begin();
    for (register long long i = 0; i < k - 1; i++, p++)
      ;
    cout << *p;
  }
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
const int INF = 0x3f3f3f3f;
const int maxn = 1e6 + 50;
const int mod = 1e9 + 7;
set<long long> s;
void divide(long long num) {
  for (register long long i = 1; i * i <= num; i++) {
    if (num % i == 0) s.insert(i), s.insert(num / i);
  }
}
int main() {
  long long n, k;
  scanf("%lld%lld", &n, &k);
  divide(n);
  if (k > s.size())
    printf("-1");
  else {
    auto p = s.begin();
    for (register long long i = 0; i < k - 1; i++, p++)
      ;
    cout << *p;
  }
}

```