Problem:
Score : 200 points

Problem Statement

You are given positive integers A and B.

Find the K-th largest positive integer that divides both A and B.

The input guarantees that there exists such a number.

Constraints

All values in input are integers.

1 \leq A, B \leq 100

The K-th largest positive integer that divides both A and B exists.

K \geq 1

Input

Input is given from Standard Input in the following format:

A B K

Output

Print the K-th largest positive integer that divides both A and B.

Sample Input 1

8 12 2

Sample Output 1

2

Three positive integers divides both 8 and 12: 1, 2 and 4.
Among them, the second largest is 2.

Sample Input 2

100 50 4

Sample Output 2

5

Sample Input 3

1 1 1

Sample Output 3

1

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.ArrayList;
import java.util.Collections;
import java.util.HashSet;
import java.util.List;
import java.util.Scanner;
import java.util.Set;

public class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int A = sc.nextInt();
    int B = sc.nextInt();
    int K = sc.nextInt();
    Main abc120B = new Main();
    System.out.println(abc120B.solve(A, B, K));
  }

  public int solve(int A, int B, int K) {
    ABC120BSolve abc120BSolve = new ABC120BSolve(A, B);
    return abc120BSolve.solve(K);
  }

  class ABC120BSolve {
    List<Integer> D;

    public ABC120BSolve(int A, int B) {
      D = new ArrayList();
      Set<Integer> divB = divisors(B);
      List<Integer> divAList = new ArrayList(divisors(A));
      Collections.sort(divAList);
      for (int a : divAList) {
        if (divB.contains(a)) {
          D.add(a);
        }
      }
    }

    public int solve(int k) {
      if (this.D.size() >= k) {
        return this.D.get(k - 1);
      } else {
        return -1;
      }

    }

    private Set<Integer> divisors(int n) {
      Set<Integer> ans = new HashSet();
      for (int i = 1; i * i <= n; i++) {
        if (n % i == 0) {
          ans.add(i);
          ans.add(n / i);
        }
      }
      return ans;
    }
  }
}