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;
bool comp(int a, int b) { return (a < b); }
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
bool isPrime(long long n) {
  if (n < 2)
    return false;
  else if (n == 2)
    return true;
  long long limit = sqrt(n);
  if (n % 2 == 0) return false;
  for (int j = 3; j <= limit; j += 2) {
    if (n % j == 0) return false;
  }
  return true;
}
int checkPalidrome(string str) {
  long long int l = str.length();
  for (int i = 0; i < l / 2; i = i + 1) {
    if (str[i] != str[l - i - 1]) return false;
  }
  return true;
}
int main() {
  long long int i, n, k = 0, t, p;
  cin >> n >> t;
  vector<long long> v;
  for (i = 1; i <= sqrt(n); i = i + 1) {
    if (n % i == 0) {
      v.push_back(i);
      if (i != sqrt(n)) v.push_back(n / i);
    }
  }
  sort(v.begin(), v.end());
  if (t > v.size())
    cout << "-1";
  else
    cout << v[t - 1];
  return 0;
}