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 = 500005;
const int SL = 100005;
const long long Mod = 1000000007;
const long long INF = 1000000000;
const long double Eps = 1e-9;
const long double PI = acos(-1.0);
const long double E = 2.7182818284590452353602874713527;
long long mnLL(long long a, long long b) { return a < b ? a : b; }
long long mxLL(long long a, long long b) { return a > b ? a : b; }
long long gcd(long long a, long long b) { return a != 0 ? gcd(b % a, a) : b; }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long sm(long long n) { return n < 10 ? n % 10 : n % 10 + sm(n / 10); }
void Sin(string& s) {
  char a[SL];
  scanf("%s", &a);
  s = a;
}
void File() {
  freopen(".in", "r", stdin);
  freopen(".out", "w", stdout);
}
long long n, k, m;
vector<long long> baj;
int main() {
  cin >> n >> k;
  for (long long i = 1; i * i <= n; i++)
    if (n % i == 0)
      if (n == i * i)
        baj.push_back(i);
      else {
        baj.push_back(i);
        baj.push_back(n / i);
      }
  sort((baj).begin(), (baj).end());
  if (baj.size() < k)
    cout << -1 << endl;
  else
    cout << baj[k - 1] << endl;
  return 0;
}