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>
#pragma comment(linker, "/STACK:66777216")
int mox[8] = {-1, -1, -1, 0, 1, 1, 1, 0};
int moy[8] = {-1, 0, 1, 1, 1, 0, -1, -1};
char mover[4] = {'D', 'U', 'R', 'L'};
const long long INF = 1LL << 61;
const int inf = 2e9;
const long double PI = acos(-1);
const long double EPS = 1e-9;
bool debug;
using namespace std;
void solve();
int main() {
  debug = 0;
  cout.precision(30);
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  srand((unsigned)time(NULL));
  solve();
  if (debug)
    cout << '\n'
         << '\n'
         << "time: " << clock() * 1.0 / CLOCKS_PER_SEC * 1000 << " ms" << '\n';
}
const int N = 1005;
const int MAGIC = 420;
const int lg = 26;
long long n, k;
vector<long long> all;
void solve() {
  cin >> n >> k;
  for (long long i = 1; i * i <= n; i++)
    if (n % i == 0) {
      all.push_back(i);
      all.push_back(n / i);
    }
  sort(all.begin(), all.end());
  all.resize(unique(all.begin(), all.end()) - all.begin());
  if ((int)all.size() <= k - 1)
    cout << -1 << '\n';
  else
    cout << all[k - 1] << '\n';
}