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;
int K, l, m;
long long N, i, j;
vector<long long> fac;
vector<pair<long long, int> > V;
void brufor(int m, long long ans) {
  if (m == -1) {
    fac.push_back(ans);
    return;
  }
  int n;
  long long kal = 1;
  for (n = 0; n <= V[m].second; n++) {
    brufor(m - 1, ans * kal);
    kal *= V[m].first;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> N >> K;
  V.push_back(pair<long long, int>(N, 1));
  for (i = 2; i <= sqrt(V[0].first); i++) {
    if (V[0].first % i == 0) {
      l++;
      V.push_back(pair<long long, int>(i, 0));
      while (V[0].first % i == 0) {
        V[0].first /= i;
        V[l].second++;
      }
    }
  }
  if (V[0].first == 1) {
    V[0].second = 0;
  }
  brufor(V.size() - 1, 1);
  if (K > fac.size()) {
    cout << -1 << endl;
    return 0;
  }
  sort(fac.begin(), fac.end());
  cout << fac[K - 1] << endl;
}