你的任务是模拟一种叫「Accordian」的纸牌游戏，他的游戏规则如下：

一副扑克牌有52张牌，首先把纸牌一张一张由左到右排好（不能有重叠，所以共有52堆牌，每堆一张），当某一张牌与他左边那张牌或者左边的第三张牌有「Match」的时候，就把这张牌移到那张牌上面去。在这里两张牌「Match」指的是这两张牌的花色（suit）或者点数（rank）一样。当你做了一个移动之后，要察看是否还可以做其他的移动。在任何时间，只有最上面那张牌可以被移动。如果因为移动一张牌使得产生一个空格（也就是被移动的那堆牌只有一张牌），你必须把右边所有的牌堆往左移一格。如此不断的寻找可移动的牌，直到没有一张牌可以移动游戏就结束了。

在选择可以移动的牌的时候可能有些状况会发生。如果有两张牌都可以移动，你应该要移动最左边的那张牌。当一张牌可以被移动到左边一格，或左边三格的时候，你必须移动到左边三格

 1# include <list>
 2# include <vector>
 3# include <cstdio>
 4# include <algorithm>
 5using namespace std;
 6list<vector<pair<char,char> > > l;
 7int main()
 8{
 9    vector<pair<char,char> >t;
10    while(true)
11    {
12        char tmp[5];
13        scanf("%s",tmp);
14        if(tmp[0]=='#') break;
15        t.clear();
16        l.clear();
17        t.push_back(pair<char,char>(tmp[0],tmp[1]));
18        l.push_back(t);
19        for(int i=1;i<52;i++)
20        {
21            t.clear();
22            scanf("%s",tmp);
23            t.push_back(pair<char,char>(tmp[0],tmp[1]));
24            l.push_back(t);
25        }
26        while(true)
27        {
28            bool flag=false;
29            for(list<vector<pair<char,char> > >::iterator i=l.begin();i!=l.end()&&!flag;)
30            {
31                int count=0;
32                bool remove=false;
33                list<vector<pair<char,char> > >::iterator i3=i;
34                while(i3!=l.begin()&&count<3)
35                {
36                    count++;
37                    i3--;
38                }
39                if(count==3&&(i3->back().first==i->back().first||i3->back().second==i->back().second))
40                {
41                    i3->push_back(i->back());
42                    i->pop_back();
43                    if(i->empty()) i=l.erase(i),remove=true;
44                    flag=true;
45                }
46                else
47                {
48                    i3=i;
49                    if(i3!=l.begin())
50                    {
51                        i3--;
52                        if(i3->back().first==i->back().first||i3->back().second==i->back().second)
53                        {
54                            i3->push_back(i->back());
55                            i->pop_back();
56                            if(i->empty()) i=l.erase(i),remove=true;
57                            flag=true;
58                        }
59                    }
60                }
61                if(!remove) i++;
62            }
63            if(!flag) break;
64        }
65        printf("%d piles remaining:",l.size());
66        for(list<vector<pair<char,char> > >::iterator i=l.begin();i!=l.end();i++)
67            printf(" %d",i->size());
68        printf("\n");
69    }
70}
71

For 2 non-negative integers x and y, f(x, y) is defined as the number of different bits in the binary format of x and y. For example, f(2, 3)=1, f(0, 3)=2, f(5, 10)=4.

Now given 2 sets of non-negative integers A and B, for each integer b in B, you should find an integer a in A such that f(a, b) is minimized. If there are more than one such integers in set A, choose the smallest one.

Input

The first line of the input is an integer T (0 < T ≤ 100), indicating the number of test cases. The first line of each test case contains 2 positive integers m and n (0 < m, n ≤ 100), indicating the numbers of integers of the 2 sets A and B, respectively. Then follow (m + n) lines, each of which contains a non-negative integers no larger than 1000000. The first m lines are the integers in set A and the other n lines are the integers in set B.

Output

For each test case you should output n lines, each of which contains the result for each query in a single line.

Sample Input

2
2 5
1
2
1
2
3
4
5
5 2
1000000
9999
1423
3421
0
13245
353

Sample Output

1
2
1
1
1
9999
0

不解释了，N2
 1# include <iostream>
 2# include <algorithm>
 3using namespace std;
 4int a[101],n,m;
 5int main()
 6{
 7    int t;
 8    cin>>t;
 9    while(t--)
10    {
11        cin>>n>>m;
12        for(int i=0;i<n;i++)
13           cin>>a[i];
14        sort(a,a+n);
15        for(int i=0;i<m;i++)
16        {
17           int t,res=0xfffffff,num;
18           cin>>t;
19           for(int j=0;j<n;j++)
20           {
21              int tmp1=a[j],tmp2=t,diff=0;
22              while(tmp1||tmp2)
23              {
24                 diff+=((tmp1&1)!=(tmp2&1));
25                 tmp1>>=1;
26                 tmp2>>=1;
27              }
28              if(diff<res) res=diff,num=a[j];
29           }
30           cout<<num<<endl;
31        }
32    }
33  //  system("pause");
34    return 0;
35}
36