題意：給你n個矩形然後在牆上按先後順序塗上顏色最後求牆上有多少顏色和該顏色對應的面積

題解：因為n很小離散化一些標記一下這塊面積是被那個顏色佔用即可，注意一些如果有多個顏色是There are並且color要加s只有一個顏色只要There is即可

#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<cstdio>
#include<cmath>
#include<set>
#include<map>
#include<cstdlib>
#include<ctime>
#include<stack>
using namespace std;
#define mes(a) memset(a,0,sizeof(a))
#define rep(i,a,b) for(i = a; i <= b; i++)
#define dec(i,a,b) for(i = b; i >= a; i--)
#define fi first
#define se second
#define ls rt<<1
#define rs rt<<1|1
#define mid (L+R)/2
#define lson ls,L,mid
#define rson rs,mid+1,R
typedef double db;
typedef long long int ll;
typedef pair<int,int> pii;
typedef unsigned long long ull;
const int mx = 1e5+5;
const int x_move[] = {1,-1,0,0,1,1,-1,-1};
const int y_move[] = {0,0,1,-1,1,-1,1,-1};
int n,m;
struct node{
    int x1,y1,x2,y2;
    int co;
}a[mx];
int x[mx],y[mx];
int vis[1005][1005];
int ans[105];
int main(){
    int t,q,ca = 1;
    while(scanf("%d%d",&n,&m),n||m){
        int d = 0,s = 0;
        scanf("%d",&n);
        mes(ans);
        mes(vis);
        for(int i = 1; i <= n; i++){
            scanf("%d%d%d%d%d",&a[i].x1,&a[i].y1,&a[i].x2,&a[i].y2,&a[i].co);
            x[++d] = a[i].x1;
            x[++d] = a[i].x2;
            y[++s] = a[i].y1;
            y[++s] = a[i].y2;
        }
        sort(x+1,x+d+1);
        sort(y+1,y+s+1);
        n = 1,m = 1;
        for(int i = 1; i <= d; i++)
            if(x[i]!=x[n])
                x[++n] = x[i];
        for(int i = 1; i <= s; i++)
            if(y[i]!=y[m])
                y[++m] = y[i];
        d/=2;
        for(int i = 1; i <= d; i++){
            int l,r,L,R;
            l = lower_bound(x+1,x+n,a[i].x1)-x;
            r = lower_bound(x+1,x+n,a[i].x2)-x;
            L = lower_bound(y+1,y+m,a[i].y1)-y;
            R = lower_bound(y+1,y+m,a[i].y2)-y;
            //cout<<l<<r<<L<<R<<endl;
            for(int j = l; j < r; j++)
                for(int k = L; k < R; k++)
                    vis[j][k] = a[i].co;
        }
        for(int i = 1; i < n; i++)
            for(int j = 1; j < m; j++){
                if(vis[i][j]){
            //        cout<<i<<j<<endl;
            //        cout<<vis[i][j]<<endl;
                    ans[vis[i][j]]+= (y[j+1]-y[j])*(x[i+1]-x[i]);
                }
            }
        int num = 0;
        if(ca>1)
            puts("");
        printf("Case %d:\n",ca++);
        for(int i = 1; i <= 100; i++)
            if(ans[i]){
                printf("%d %d\n",i,ans[i]);
                num++;
            }
        if(num==1)
            printf("There is %d color left on the wall.\n",num);
        else
            printf("There are %d colors left on the wall.\n",num);
    }
    return 0;
}