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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

32 3 12 3 28 16 3

Sample Output

1-12

题意：

给X,Y,K求X,Y的第k大的因子。

思路：

第k的大因子一定是X，Y的最大公因子的因子，所以我们可以先求出X，Y的因子，然后暴力枚举每一个因子（O(sqrt(n))的复杂度），如果数量比k大，排序找出，否则输出-1。

上板子，今天刚找的。

#include
   
    #include
    
     #include
     
      using namespace std;typedef long long ll;ll ans[2000100];ll gcd(ll a,ll b){	if(b==0)return a;	return gcd(b,a%b);}ll cmp(ll a,ll b){	return a>b;}void solve(ll x,ll k){	int cnt=0;	for(ll i=1;i*i<=x;i++)	{		if(x%i==0)		{			ans[cnt++]=i;			if(x/i!=i)			ans[cnt++]=x/i;		}	}	if(cnt>=k)	{		sort(ans,ans+cnt,cmp);		cout<
      
       <
       
        >t;	while(t--)	{		ll x,y,k;		cin>>x>>y>>k;		ll g=gcd(x,y);		solve(g,k);	}	return 0;}
       
      
     
    
   

 

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