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 GCC optimize("O2")
using namespace std;
inline long long int input() {
  long long int n;
  cin >> n;
  return n;
}
long long int pw(long long int a, long long int b) {
  return (!b ? 1 : (b & 1 ? a * pw(a * a, b / 2) : pw(a * a, b / 2)));
}
const long long int MAXN = 2e5 + 10;
const long long int MAXA = 101;
const long long int INF = 8e18;
const long long int MOD = 1e9 + 7;
const long double PI = 4 * atan((long double)1);
vector<long long int> v, ans;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
  ;
  long long int n, k;
  cin >> n >> k;
  for (long long int i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      v.push_back(i);
      if (i * i != n) ans.push_back(n / i);
    }
  }
  reverse((ans).begin(), (ans).end());
  if (k > (long long int)v.size() + (long long int)ans.size())
    return cout << -1 << endl, 0;
  ;
  if (k <= (long long int)v.size()) return cout << v[k - 1] << endl, 0;
  ;
  return cout << ans[k - (long long int)v.size() - 1] << endl, 0;
  ;
  return 0;
}