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;
const long long modu = 1e9 + 7;
int noofdig(int N) { return floor(log10(N)) + 1; }
int bits_count(unsigned int u) {
  unsigned int uCount;
  uCount = u - ((u >> 1) & 033333333333) - ((u >> 2) & 011111111111);
  return ((uCount + (uCount >> 3)) & 030707070707) % 63;
}
void solve() {
  long long int n, k;
  cin >> n >> k;
  vector<long long int> div;
  for (long long int i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      div.push_back(i);
      if (i * i != n) div.push_back(n / i);
    }
  }
  sort(div.begin(), div.end());
  if ((long long int)div.size() >= k)
    cout << div[k - 1] << '\n';
  else
    cout << -1 << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  srand(chrono::high_resolution_clock::now().time_since_epoch().count());
  int t = 1;
  while (t--) {
    solve();
  }
}