#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#define maxn 205
#define MAXN 100005
#define mod 100000000
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-8
typedef long long ll;
using namespace std;

ll n,m,ans,cnt,tot,flag;
ll mp[12][12],dp[12][12][1<<12];

void solve()
{
    ll i,j,k,t;
    ll up,down,left,right;
    tot=(1<<m+1)-1;
    memset(dp,0,sizeof(dp));
    dp[0][m][0]=1;
    for(i=1;i<=n;i++)
    {
        for(k=0;k<(1<<m);k++) // 行与行之间单独转移
        {
            dp[i][0][k<<1]=dp[i-1][m][k];
        }
        for(j=1;j<=m;j++)
        {
            for(k=0;k<=tot;k++) // i j-1的状态
            {
                if(dp[i][j-1][k]==0) continue ;
                down=left=1<<j-1;
                right=up=1<<j;
                if(mp[i][j])
                {
                    if((k&left)&&(k&up))  // 两个插头
                    {
                        dp[i][j][k^left^up]+=dp[i][j-1][k];
                    }
                    else if(!(k&left)&&!(k&up)) // 0个插头
                    {
                        dp[i][j][k^left^up]+=dp[i][j-1][k];
                    }
                    else // 一个插头
                    {
                        dp[i][j][k]+=dp[i][j-1][k];
                        dp[i][j][k^left^up]+=dp[i][j-1][k];
                    }
                }
                else
                {
                    if(!(k&left)&&!(k&up)) dp[i][j][k]+=dp[i][j-1][k];
                }
            }
        }
    }
    ans=dp[n][m][0];
}
int main()
{
    ll i,j,t,test=0;
    scanf("%I64d",&t);
    while(t--)
    {
        scanf("%I64d%I64d",&n,&m);
        for(i=1;i<=n;i++)
        {
            for(j=1;j<=m;j++)
            {
                scanf("%I64d",&mp[i][j]);
            }
        }
        solve();
        printf("Case %I64d: There are %I64d ways to eat the trees.\n",++test,ans);
    }
    return 0;
}