#pragma comment(linker, "/STACK:102400000, 102400000")

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<cctype>

#include<cmath>

#include<iostream>

#include<sstream>

#include<iterator>

#include<algorithm>

#include<string>

#include<vector>

#include<set>

#include<map>

#include<stack>

#include<deque>

#include<queue>

#include<list>

#define Min(a, b) ((a < b) ? a : b)

#define Max(a, b) ((a < b) ? b : a)

typedef long long ll;

typedef unsigned long long llu;

const int INT_INF = 0x3f3f3f3f;

const int INT_M_INF = 0x7f7f7f7f;

const ll LL_INF = 0x3f3f3f3f3f3f3f3f;

const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;

const int dr[] = {0, -1, 0, 1, -1, -1, 1, 1};//西北东南

const int dc[] = {-1, 0, 1, 0, -1, 1, -1, 1};

const int MOD = 1e9 + 7;

const double pi = acos(-1.0);

const double eps = 1e-15;

const int MAXN = 10 + 10;

const int MAXT = 10000 + 10;

using namespace std;

char pic[MAXN][MAXN];

char p1[MAXN][MAXN];

int vis[MAXN];

bool judge(){//判断是否摆成目标形状

    for(int i = 0; i < 5; ++i){

        for(int j = 0; j < 9; ++j){

            if(pic[i][j] != p1[i][j]) return false;

        }

    }

    return true;

}

bool dfs(int step){

    if(judge()) return true;

    if(step >= 6) return false;

    char p2[MAXN][MAXN];

    memcpy(p2, p1, sizeof p1);//p2便于恢复p1数组内容

    for(int i = 0; i < 9; ++i){//枚举2*2正方形可以放置的9个位置

        if(!vis[i]){

            vis[i] = 1;

            int r = i / 3;

            int c = 2 * (i % 3) + 1;

            p1[r][c] = p1[r][c + 2] = p1[r + 2][c] = p1[r + 2][c + 2] = '_';

            p1[r + 1][c - 1] = p1[r + 2][c - 1] = p1[r + 1][c + 3] = p1[r + 2][c + 3] = '|';

            p1[r + 1][c] = p1[r + 1][c + 1] = p1[r + 1][c + 2] = p1[r + 2][c + 1] = ' ';

            if(dfs(step + 1)) return true;

            vis[i] = 0;

            memcpy(p1, p2, sizeof p1);

        }

    }

    return false;

}

int main(){

    int kase = 0;

    while(gets(pic[0]) != NULL){

        if(pic[0][0] == '0') return 0;

        memset(vis, 0, sizeof vis);

        for(int i = 1; i < 5; ++i){

            gets(pic[i]);

        }

        printf("Case %d: ", ++kase);

        for(int i = 0; i < 5; ++i){

            for(int j = 0; j < 9; ++j){

                p1[i][j] = ' ';

            }

        }

        if(dfs(0)) printf("Yes\n");

        else printf("No\n");

    }

    return 0;

}