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 si, 1, si, 2, ... , si, mi (1 ≤ |si, j| ≤ 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 Friends need to be identified, and output "Let's go to the library!!" (without quotes) on the third
line.

题意：

就是给你几个人的名字，每组包含几个名字，如果含有自己的名字，就为1，否则为0,根据几组数据，能否判断他叫什么名字！

思路细节：数据要足够大，采用map,记录名字序号，

代码：

#include<bits/stdc++.h>
using namespace std;
int main()
{
//freopen("aa.txt","r",stdin);
int t,n,m,q,c,i,j,k;
cin>>t;
while(t--)
{
cin>>n>>q;
cin>>c;
bool b[300][300]={0};

map<string,int>name;
string s[300];
string ss;
for(i=1;i<=c;i++)
{
cin>>s[i];
name[s[i]]=i;
}
for(i=1;i<=q;i++)
{
cin>>m;
for(j=1;j<=m;j++)
{
cin>>ss;
b[i][name[ss]]=1;
}
}
bool bb;
for(i=1;i<=n;i++)
{
int a[300]={0};
for(j=1;j<=q;j++)
{
cin>>bb;
for(k=1;k<=c;k++)
{
if(b[j][k]!=bb)
a[k]--;
}
}
int xx=0,yy=0;
for(k=1;k<=c;k++)
{
if(a[k]==0)
{
xx=k;
yy++;
}
}
if(yy==0||yy>1)
cout<<"Let's go to the library!!"<<endl;
else
cout<<s[xx]<<endl;
}

}

return 0;
}