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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

代码及注释如下：

/*题意：求两个数a,b公共的从大到小的第k个因子...先求出a,b的Gcd后枚举求Gcd的因子....为啥呢...因为a,b都能整除Gcd,所以a,b肯定能够整除Gcd的因子...用的优先队列储存因子....注意数据要用long long int .....*/#include 
   
    #include 
    
     #include 
     
      #include 
      
       #include 
       
        using namespace std;int t;long long int x,y,k;priority_queue 
        
         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()
        
       
      
     
    
   

 

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