#include <bits/stdc++.h>
using namespace std;
int main() {
  long long i, n, k, sum = 0;
  bool flag = false;
  set<long long> S;
  scanf("%I64d%I64d", &n, &k);
  for (i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      S.insert(i);
      S.insert(n / i);
    }
  }
  if (S.size() >= k) flag = true;
  if (flag) {
    set<long long>::iterator it;
    it = S.begin();
    while (k--) {
      it++;
    }
    it--;
    cout << *it;
  } else
    cout << -1;
  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;
int main() {
  long long i, n, k, sum = 0;
  bool flag = false;
  set<long long> S;
  scanf("%I64d%I64d", &n, &k);
  for (i = 1; i <= sqrt(n); i++) {
    if (n % i == 0) {
      S.insert(i);
      S.insert(n / i);
    }
  }
  if (S.size() >= k) flag = true;
  if (flag) {
    set<long long>::iterator it;
    it = S.begin();
    while (k--) {
      it++;
    }
    it--;
    cout << *it;
  } else
    cout << -1;
  return 0;
}

```