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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
map<unsigned long long, int> factors;
vector<unsigned long long> res;
void generate(map<unsigned long long, int>::iterator it,
              unsigned long long numb) {
  if (it == factors.end()) {
    res.push_back(numb);
    return;
  }
  unsigned long long pown = 1;
  generate(++it, numb * pown);
  --it;
  for (int i = 0; i < it->second; ++i) {
    pown *= it->first;
    generate(++it, numb * pown);
    --it;
  }
}
int main() {
  unsigned long long n, k;
  cin >> n >> k;
  unsigned long long val = n;
  bool out = true;
  for (unsigned long long i = 2; i <= sqrt(val); ++i) {
    if (val % i == 0) {
      factors[i]++;
      val /= i;
      i = 1;
    }
  }
  if (val != 1) factors[val]++;
  generate(factors.begin(), 1);
  sort(res.begin(), res.end());
  auto end = unique(res.begin(), res.end());
  int size = distance(res.begin(), end);
  if (k > size) {
    cout << -1 << endl;
  } else {
    cout << res[k - 1] << endl;
  }
}