Problem Description

Alice has received a beautiful present from Bob. The present contains n lanterns and m switches. Each switch controls some lanterns and pushing the switch will change the state of all lanterns it controls from off to on or from on to off. A lantern may be controlled by many switches. At the beginning, all the lanterns are off.

Alice wants to change the state of the lanterns to some specific configurations and she knows that pushing a switch more than once is pointless. Help Alice to find out the number of ways she can achieve the goal. Two ways are different if and only if the sets (including the empty set) of the switches been pushed are different.

 

Input

The first line contains an integer T (T<=5) indicating the number of test cases.
The first line of each test case contains an integer n (1<=n<=50) and m (1<=m<=50).
Then m lines follow. Each line contains an integer k (k<=n) indicating the number of lanterns this switch controls.
Then k integers follow between 1 and n inclusive indicating the lantern controlled by this switch.
The next line contains an integer Q (1<=Q<=1000) represent the number of queries of this test case.
Q lines follows. Each line contains n integers and the i-th integer indicating that the state (1 for on and 0 for off) of the i-th lantern of this query.

 

Output

For each test case, print the case number in the first line. Then output one line containing the answer for each query.
Please follow the format of the sample output.

 

Sample Input

 

Sample Output

 

Source

 //#include"bits/stdc++.h"

 #include<sstream>

 #include<iomanip>

 #include"cstdio"

 #include"map"

 #include"set"

 #include"cmath"

 #include"queue"

 #include"vector"

 #include"string"

 #include"cstring"

 #include"time.h"

 #include"iostream"

 #include"stdlib.h"

 #include"algorithm"

 #define db double

 #define ll long long

 #define vec vector<ll>

 #define mt  vector<vec>

 #define ci(x) scanf("%d",&x)

 #define cd(x) scanf("%lf",&x)

 #define cl(x) scanf("%lld",&x)

 #define pi(x) printf("%d\n",x)

 #define pd(x) printf("%f\n",x)

 #define pl(x) printf("%lld\n",x)

 //#define rep(i, x, y) for(int i=x;i<=y;i++)

 #define rep(i, n) for(int i=0;i<n;i++)

 const int N   = 1e2+ ;

 //const int mod = 1e9 + 7;

 //const int MOD = mod - 1;

 const int inf = 0x3f3f3f3f;

 const db  PI  = acos(-1.0);

 const db  eps = 1e-;

 using namespace std;

 int equ,var;//equ个方程，var个变量，增广矩阵行数为equ,列数为var+1,分别为0到var

 int a[N][N];//增广矩阵

 int x[N];//存储自由变元

 int f_x[N];

 int free_x;//自由变元个数

 void swap(int &x,int &y){

     int t;

     t=x,x=y,y=t;

 }

 int Gauss()

 {

     int ma_r,col,k;

     free_x=;

     for(k=,col=;k<equ&&col<var;k++,col++){

         ma_r = k;

         for (int i = k + ; i < equ; i++) if (abs(a[i][col] > abs(a[ma_r][col]))) ma_r = i;//取系数最大的一行

         if (!a[ma_r][col]) {

             k--;

             f_x[free_x++] = col;

             continue;

         }

         if (ma_r != k)

             for (int j = col; j < var + ; j++) swap(a[k][j], a[ma_r][j]);//与当前行交换

         for (int i = k + ; i < equ; i++)

             if (a[i][col] != )

                 for (int j = col; j < var + ; j++) a[i][j] ^= a[k][j];//消除其他行第col列的变量

     }

     for(int i=k;i<equ;i++) if(a[i][col]!=) return -;//没被消除则无解

     if(k<var) return var-k;//自由变元个数

     //唯一解，回代

     for(int i=var-;i>=;i--){

         x[i]=a[i][var];

         for(int j=i+;j<var;j++) x[i]^=(a[i][j]&&x[j]);//自下而上

     }

     return ;

 }

 int b[N][N];

 int n,m;

 int main()

 {

     int t;

     ci(t);

     for(int I=;I<=t;I++)

     {

         ci(n),ci(m);

         memset(a,, sizeof(a));

         for(int i=;i<m;i++){

             int k,c;

             ci(k);

             for(int j=;j<k;j++) ci(c),a[c-][i]=;

         }

         equ=n,var=m;

         for(int i=;i<equ;i++){

             for(int j=;j<var;j++){

                 b[i][j]=a[i][j];

             }

         }

         int q;

         ci(q);

         printf("Case %d:\n",I);

         while(q--)

         {

             for(int i=;i<equ;i++)

                 for(int j=;j<var;j++)

                     a[i][j]=b[i][j];

             for(int i=;i<n;i++) ci(a[i][var]);

             int ans=Gauss();

             if(ans==-) puts("");

             else pl(1ll<<ans);

         }

     }

     return ;

 }