Now I am leaving hust acm. In the past two and half years, I learned so many knowledge about Algorithm and Programming, and I met so many good friends. I want to say sorry to Mr, Yin, I must leave now ~~>.<~~. I am very sorry, we could not advanced to the World Finals last year.
When coming into our training room, a lot of books are in my eyes. And every time the books are moving from one place to another one. Now give you the position of the books at the early of the day. And the moving information of the books the day, your work is to tell me how many books are stayed in some rectangles.
To make the problem easier, we divide the room into different grids and a book can only stayed in one grid. The length and the width of the room are less than 1000. I can move one book from one position to another position, take away one book from a position or bring in one book and put it on one position.

Input

In the first line of the input file there is an Integer T(1<=T<=10), which means the number of test cases in the input file. Then N test cases are followed.
For each test case, in the first line there is an Integer Q(1<Q<=100,000), means the queries of the case. Then followed by Q queries.
There are 4 kind of queries, sum, add, delete and move.
For example:
S x1 y1 x2 y2 means you should tell me the total books of the rectangle used (x1,y1)-(x2,y2) as the diagonal, including the two points.
A x1 y1 n1 means I put n1 books on the position (x1,y1)
D x1 y1 n1 means I move away n1 books on the position (x1,y1), if less than n1 books at that position, move away all of them.
M x1 y1 x2 y2 n1 means you move n1 books from (x1,y1) to (x2,y2), if less than n1 books at that position, move away all of them.
Make sure that at first, there is one book on every grid and 0<=x1,y1,x2,y2<=1000,1<=n1<=100.

Output

At the beginning of each case, output "Case X:" where X is the index of the test case, then followed by the "S" queries.
For each "S" query, just print out the total number of books in that area.

Sample Input

2
3
S 1 1 1 1
A 1 1 2
S 1 1 1 1
3
S 1 1 1 1
A 1 1 2
S 1 1 1 2

Sample Output

Case 1:
1
3
Case 2:
1
4

題意：把一個房間劃分成一個矩陣，其中的每個放個都可以放書，且每個方格初始都有一本書，接下來有四種操作：S x1 y1 x2 y2是詢問a[x1][y1]到a[x2][y2]直接有多少本書，A x1 y1 n1意思是向a[x1][y1]中放n1本書，D x1 y1 n1意思是從a[x1][y1]中拿出n1本書，如果不足則全部拿出。M x1 y1 x2 y2 n1意為從a[x1][y1]中拿出n1本書放到a[x2][y2]中，若不足則全部拿出。

這應該是一道比較簡單的二維樹狀陣列的題，詳細請看程式碼：

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;
const int N=1005;
int a[N][N],c[N][N];
int lowbit(int i)
{
    return i&(-i);
}
void update(int x,int y,int val)
{
    int i=y;
    while(x<=1001)//這裡最小要開到1001，因為開始時會將x和y++，這裡哇了很久啊！！！
    {
        y=i;
        while(y<=1001)
        {
            c[x][y]+=val;
            y+=lowbit(y);
        }
        x+=lowbit(x);
    }
}
int sum(int x,int y)
{
    int ans=0,i=y;
    while(x>0)
    {
        y=i;
        while(y>0)
        {
            ans+=c[x][y];
            y-=lowbit(y);
        }
        x-=lowbit(x);
    }
    return ans;
}
int main()
{
    char s[2];
    int n,t,k=1,m,x1,x2,y1,y2;
    scanf("%d",&t);
    while(t--)
    {
        memset(a,0,sizeof(a));
        memset(c,0,sizeof(c));
        for(int i=1;i<=1001;i++)//這裡也必須到1001，你懂得
        {
            for(int j=1;j<=1001;j++)
            {
                a[i][j]=1;
                update(i,j,1);
            }
        }
        printf("Case %d:\n",k++);
        scanf("%d",&m);
        for(int i=0;i<m;i++)
        {
            scanf("%s",s);
            if(s[0]=='S')
            {
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                x1++,x2++,y1++,y2++;
                if(x1>x2) swap(x1,x2);
                if(y1>y2) swap(y1,y2);
                //二維求得是一個矩陣的和，劃一下圖就知道為什麼要這樣寫了，用總的減去左邊和右邊不要的，然後再加上多減的一個矩陣
                printf("%d\n",sum(x2,y2)-sum(x2,y1-1)-sum(x1-1,y2)+sum(x1-1,y1-1));
            }
            else if(s[0]=='A')
            {
                scanf("%d%d%d",&x1,&y1,&n);
                x1++,y1++;
                a[x1][y1]+=n;
                update(x1,y1,n);
            }
            else if(s[0]=='D')
            {
                scanf("%d%d%d",&x1,&y1,&n);
                x1++,y1++;
                if(a[x1][y1]<n) n=a[x1][y1];
                update(x1,y1,-n);
                a[x1][y1]-=n;//
            }
            else
            {
                scanf("%d%d%d%d%d",&x1,&y1,&x2,&y2,&n);
                x1++,y1++,x2++,y2++;
                if(a[x1][y1]<n) n=a[x1][y1];
                update(x1,y1,-n);
                update(x2,y2,n);
                a[x1][y1]-=n;//
                a[x2][y2]+=n;//
            }
        }
    }
}