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 int N = 1e5 + 10;
const long long inf = 1e2;
map<long long, bool> mp;
vector<long long> vc, p;
void insert(long long n, int len) {
  long long in;
  for (int i = 0; i < len; ++i) {
    in = vc[i] * n;
    vc.push_back(in);
  }
}
void prime(long long n) {
  long long nn = n, div;
  int len;
  for (long long i = 2; i * i <= nn; ++i) {
    if (n % i == 0) {
      len = vc.size();
      div = 1;
      while (n % i == 0) {
        n /= i;
        div *= i;
        insert(div, len);
      }
    }
  }
  if (n != 1) {
    insert(n, vc.size());
  }
}
int main() {
  long long n, k;
  cin >> n >> k;
  vc.push_back(1);
  prime(n);
  sort(vc.begin(), vc.end());
  if (vc.size() < k)
    cout << -1 << endl;
  else
    cout << vc[k - 1] << endl;
}