#include <iostream>
using namespace std;
typedef __int64 LL;
LL dp[12][12][1<<12];
long hash[12][12];

int main()
{
    long T;
    long b=1;
    scanf("%ld",&T);
    while (T--)
    {
        long i,j,k,m,n;
        scanf("%ld %ld",&m,&n);
        for (i=1;i<=m;++i)
        {
            for (j=1;j<=n;++j)
            {
                scanf("%ld",&hash[i][j]);
            }
        }

dp[0][n][0]=1;

for (i=1;i<=m;++i)
        {
            long len=1<<n;
            for (j=0;j<len;++j)
            {
                dp[i][0][j<<1]=dp[i-1][n][j];
            }

for (j=1;j<=n;++j)
            {
                len=(1<<n<<1);
                for(k=0;k<len;++k)
                {
                    long p=1<<j;
                    long q=p>>1;
                    
                    bool x=p&k;
                    bool y=q&k;

if (hash[i][j])
                    {
                        dp[i][j][k]=dp[i][j-1][k^p^q];
                        if (x!=y)
                        {
                            dp[i][j][k]+=dp[i][j-1][k];
                        }
                    }
                    else
                    {
                        if (x==0&&y==0)
                        {
                            dp[i][j][k]=dp[i][j-1][k];
                        }
                        else
                        {
                            dp[i][j][k]=0;
                        }
                    }
                }
            }
        }
        printf("Case %ld: There are %I64d ways to eat the trees.\n",b++,dp[m][n][0]);

}
    return 0;
}

#include<iostream>
using namespace std;
int mat[100][100];
__int64 dp[12][12][1<<12];
int main()
{
    // freopen("1.txt","r",stdin);
    int zu;
    int g=1;
    scanf("%d",&zu);
    while(zu--)
    {
        int m,n;
        scanf("%d%d",&m,&n);
        int i,j,k;
        for(i=1;i<=m;i++)
            for(j=1;j<=n;j++)
                scanf("%d",&mat[i][j]);
            //memset(dp,0,sizeof(dp));
            dp[0][n][0]=1;
            for(i=1;i<=m;i++)
            {
                for(j=0;j<(1<<n);j++)
                    dp[i][0][j<<1]=dp[i-1][n][j];
                //把最右边的去掉 把所有的状态拉到下一行并进行合理的改变(注意最左边的是低位所代表的数) 
                //上一行最右边的边格一定不与轮廓线重合 下一行最左边的边格也一定不与轮廓线重合 
                //所以进行的是上一行所有状态(也就是轮廓线)向右的一次滑动
                    for(j=1;j<=n;j++)//枚举决策线的拐向
                    {
                        for(k=0;k<(1<<n<<1);k++)//枚举轮廓线
                        {
                            int p=1<<j;//第j个轮廓段(上)
                            int q=p>>1;//第j-1个轮廓段(左)
                            bool x=k&p;//左轮廓是否与通路相交
                            bool y=k&q;//上轮廓是否与通路相交
                            
                            //判断轮廓线在[i,j]为拐向时的分布 每个格子有1<<n<<1种 多阶段决策
                            
                            if(mat[i][j])//如果该单元可以通行
                            {
                                dp[i][j][k]=dp[i][j-1][k^p^q];//必然有一个通路连接上一个格的通路
                                if(x!=y)
                                    dp[i][j][k]+=dp[i][j-1][k];//有一处新覆盖则有另外的一种情况
                            }
                            else//否则为障碍格子
                            {
                                if(x==0&&y==0)//通路与轮廓线不相交
                                    dp[i][j][k]=dp[i][j-1][k];//直接转移
                                else
                                    dp[i][j][k]=0;//通路与轮廓线相交的方案一定为0
                            }
                        }
                    }
            }
            printf("Case %d: There are %I64d ways to eat the trees.\n",g++,dp[m][n][0]);
    }
    return 0;
}