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;
using ll = long long int;
using pii = pair<int, int>;
using pid = pair<int, double>;
using pLL = pair<ll, ll>;
using pll = pair<long int, long int>;
using pis = pair<int, string>;
using vL = vector<ll>;
using vi = vector<int>;
using vs = vector<string>;
using vd = vector<double>;
using vb = vector<bool>;
const double pi = acos(-1);
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
bool comp(const pii &a, const pii &b) { return a.second < b.second; }
void solve() {
  ll n, k;
  cin >> n >> k;
  vL div;
  for (ll i = 1; i * i <= n; i++) {
    if ((n % i) == 0) {
      div.push_back(i);
      if ((n / i) != i) div.push_back(n / i);
    }
  }
  sort(div.begin(), div.end());
  ;
  cout << (k > div.size() ? -1 : div[k - 1]) << endl;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  solve();
  return 0;
}