What Kind of Friends Are You?

Japari Park is a large zoo home to extant species, endangered species, extinct species, cryptids and some legendary creatures. Due to a mysterious substance known as Sandstar, all the animals have become anthropomorphized into girls known as Friends.

Kaban is a young girl who finds herself in Japari Park with no memory of who she was or where she came from. Shy yet resourceful, she travels through Japari Park along with Serval to find out her identity while encountering more Friends along the way, and eventually discovers that she is a human.

However, Kaban soon finds that it's also important to identify other Friends. Her friend, Serval, enlightens Kaban that she can use some questions whose expected answers are either "yes" or "no" to identitfy a kind of Friends.

To be more specific, there are n Friends need to be identified. Kaban will ask each of them q same questions and collect their answers. For each question, she also gets a full list of animals' names that will give a "yes" answer to that question (and those animals who are not in the list will give a "no" answer to that question), so it's possible to determine the name of a Friends by combining the answers and the lists together.

But the work is too heavy for Kaban. Can you help her to finish it?

Input

There are multiple test cases. The first line of the input is an integer T (1 ≤ T≤ 100), indicating the number of test cases. Then T test cases follow.

The first line of each test case contains two integers n (1 ≤ n ≤ 100) and q (1 ≤ q ≤ 21), indicating the number of Friends need to be identified and the number of questions.

The next line contains an integer c (1 ≤ c ≤ 200) followed by c strings p1, p2, ... , pc (1 ≤ |pi| ≤ 20), indicating all known names of Friends.

For the next q lines, the i-th line contains an integer mi (0 ≤ mi ≤ c) followed by mi strings si1, si2, ... , simi (1 ≤ |sij| ≤ 20), indicating the number of Friends and their names, who will give a "yes" answer to the i-th question. It's guaranteed that all the names appear in the known names of Friends.

For the following n lines, the i-th line contains q integers ai, 1, ai, 2, ... , ai, q(0 ≤ ai, j ≤ 1), indicating the answer (0 means "no", and 1 means "yes") to the j-th question given by the i-th Friends need to be identified.

It's guaranteed that all the names in the input consist of only uppercase and lowercase English letters.

Output

For each test case output n lines. If Kaban can determine the name of the i-th Friends need to be identified, print the name on the i-th line. Otherwise, print "Let's go to the library!!" (without quotes) on the i-th line instead.

Sample Input

2
3 4
5 Serval Raccoon Fennec Alpaca Moose
4 Serval Raccoon Alpaca Moose
1 Serval
1 Fennec
1 Serval
1 1 0 1
0 0 0 0
1 0 0 0
5 5
11 A B C D E F G H I J K
3 A B K
4 A B D E
5 A B K D E
10 A B K D E F G H I J
4 B D E K
0 0 1 1 1
1 0 1 0 1
1 1 1 1 1
0 0 1 0 1
1 0 1 1 1

Sample Output

Serval
Let's go to the library!!
Let's go to the library!!
Let's go to the library!!
Let's go to the library!!
B
Let's go to the library!!
K

Hint

The explanation for the first sample test case is given as follows:

As Serval is the only known animal who gives a "yes" answer to the 1st, 2nd and 4th question, and gives a "no" answer to the 3rd question, we output "Serval" (without quotes) on the first line.

As no animal is known to give a "no" answer to all the questions, we output "Let's go to the library!!" (without quotes) on the second line.

Both Alpaca and Moose give a "yes" answer to the 1st question, and a "no" answer to the 2nd, 3rd and 4th question. So we can't determine the name of the third Friendsneed to be identified, and output "Let's go to the library!!" (without quotes) on the third line.

---

题意： n表示需要识别的朋友，q表示问题数，

c表示所有已知朋友的数量，接下来是所有已知朋友的名字。

接下来q行，第i行代表第i个问题，每一行的第一个数代表对第i个问题做出“是”回答的人数，后面为回答“是”的朋友的名字。

接下来n行，第i行第j个表示需要识别第i个朋友给出的第j个问题的答案(0表示“否”，1表示“是”)

输出：输出n行。如果能够确定需要识别的第i个朋友的名字，那么就在第i行上打印这个名字。否则，“Let's go to the library!!”

---

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<queue>
#include<cmath>
#include<stack>
using namespace std;
char name[220][30];
int vis[200];
int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        string value[200];
        int n, q;
        scanf("%d%d", &n, &q);
        int x;
        scanf("%d", &x);

        //输入所有朋友的名字；
        for(int i=0; i<x; i++)
        {
            scanf("%s", name[i]);
        }

        //循环q个问题；
        for(int i=0; i<q; i++)
        {
            int m;
            scanf("%d", &m);
            memset(vis, 0, sizeof(vis));
            for(int j=0; j<m; j++)
            {
                char s[30];
                scanf("%s", s);//对第j个问题，回答“是”的朋友的名字；

                //与所有朋友的名字比较，如果与第h个朋友名字一样，value[h]+'1'得出每个朋友答“是”的，并且标记该朋友； 
                for(int h=0; h<x; h++)
                {
                    if(!strcmp(s, name[h]))//字符串相等，则strcmp()值为0；
                    {
                        value[h]+='1';
                        vis[h] = 1;
                    }
                }

            }
            //如果朋友没有在回答该问题的“是”里面，value[j]+'0';(注意这里同上面一样都是加'0/1'直接连在字符串后面，而不是直接加)；
            for(int j=0; j<x; j++)
            {
                if(!vis[j])
                    value[j]+='0';
            }
        }

        //输入需要识别的n个朋友对每个问题的“是”、“否”；
        for(int i=0; i<n; i++)
        {
            string a;
            //对每一个需要被识别朋友的答案加和；
            for(int j=0; j<q; j++)
            {
                char x;
                scanf("%s", &x);
                a+=x;
            }
            int cnt = 0, k = 0;
            //循环x个朋友，如果有与其相等的加和且唯一，则输出朋友的名字；
            for(int j=0; j<x; j++)
            {
                if(a==value[j])
                {
                    cnt++;
                    k = j;
                }
            }
            if(cnt==1)
            {
                printf("%s\n", name[k]);
            }
            else
                printf("Let's go to the library!!\n");
        }
    }
    return 0;
}