Problem Description

In mathematics, the greatest common divisor (gcd), also known as the greatest common factor (gcf), highest common factor (hcf), or greatest common measure (gcm), of two or more integers (when at least one of them is not zero), is the largest positive integer that divides the numbers without a remainder.
---Wikipedia

Today, GCD takes revenge on you. You have to figure out the k-th GCD of X and Y.

Input

The first line contains a single integer T, indicating the number of test cases.

Each test case only contains three integers X, Y and K.

[Technical Specification]
1. 1 <= T <= 100
2. 1 <= X, Y, K <= 1 000 000 000 000

Output

For each test case, output the k-th GCD of X and Y. If no such integer exists, output -1.

Sample Input

3

2 3 1

2 3 2

8 16 3

Sample Output

1

-1

2

程式碼及註釋如下：

/*
題意：
求兩個數a,b公共的從大到小的第k個因子...
先求出a,b的Gcd後列舉求Gcd的因子....
為啥呢...
因為a,b都能整除Gcd,所以a,b肯定能夠整除Gcd的因子...
用的優先佇列儲存因子....
注意資料要用long long int .....
*/
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>
using namespace std;
int t;
long long int x,y,k;
priority_queue <long long int>q;
long long int gcd (long long int a,long long int b)
{
    return b? gcd(b,a%b):a;
}
void Clear()
{
    while (!q.empty())
        q.pop();
}
int main()
{
    scanf("%d",&t);
    while (t--)
    {
        Clear();
        scanf("%lld%lld%lld",&x,&y,&k);
        long long int Gcd=gcd(x,y);
        q.push(Gcd);
        if(Gcd!=1)
        {
            q.push(1);
           for (long long int i=2;i*i<=Gcd;i++)
           {
               if(Gcd%i==0)
               {
                   q.push(i);
                   if(i*i!=Gcd)
                   {
                      q.push(Gcd/i);
                   }
               }
           }
        }
        if(q.size()<k)
            printf("-1\n");
        else
        {
            for (int i=0;i<k-1;i++)
                {
                    q.pop();
                }
            printf("%lld\n",q.top());
        }

    }
    return 0;
}