Matrix

Given an N*N matrix A, whose elements are either 0 or 1. A[i, j] means the number in the i-th row and j-th column. Initially we have A[i, j] = 0 (1 <= i, j <= N). 

We can change the matrix in the following way. Given a rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2), we change all the elements in the rectangle by using "not" operation (if it is a '0' then change it into '1' otherwise change it into '0'). To maintain the information of the matrix, you are asked to write a program to receive and execute two kinds of instructions. 

1. C x1 y1 x2 y2 (1 <= x1 <= x2 <= n, 1 <= y1 <= y2 <= n) changes the matrix by using the rectangle whose upper-left corner is (x1, y1) and lower-right corner is (x2, y2). 
2. Q x y (1 <= x, y <= n) querys A[x, y]. 

Input

The first line of the input is an integer X (X <= 10) representing the number of test cases. The following X blocks each represents a test case. 

The first line of each block contains two numbers N and T (2 <= N <= 1000, 1 <= T <= 50000) representing the size of the matrix and the number of the instructions. The following T lines each represents an instruction having the format "Q x y" or "C x1 y1 x2 y2", which has been described above. 

Output

For each querying output one line, which has an integer representing A[x, y]. 

There is a blank line between every two continuous test cases. 

Sample Input

1
2 10
C 2 1 2 2
Q 2 2
C 2 1 2 1
Q 1 1
C 1 1 2 1
C 1 2 1 2
C 1 1 2 2
Q 1 1
C 1 1 2 1
Q 2 1

Sample Output

1
0
0
1
#include<iostream>
#include<stdio.h>
using namespace std;
const int Maxn=1010;
int n,m,c[Maxn][Maxn];
int lowbit(int x){
    return x&(-x);
}
void updata(int x,int y,int val){
    for(int i=x;i<=n;i+=lowbit(i))
        for(int j=y;j<=n;j+=lowbit(j))
            c[i][j]+=val;
}
int getSum(int x,int y){
    int ret=0;
    for(int i=x;i>0;i-=lowbit(i))
        for(int j=y;j>0;j-=lowbit(j))
            ret+=c[i][j];
    return ret;
}
int main(){
    //freopen("in.txt","r",stdin);
    int t;scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++)
                c[i][j]=0;
        char s;int x1,y1,x2,y2;
        while(m--){
            cin>>s;
            if(s=='C'){
                scanf("%d%d%d%d",&x1,&y1,&x2,&y2);
                //目标是(x1,y1)~(x2,y2)矩阵整体加1
                updata(x1,y1,1);//(x1,y1)~(n,n)矩阵整体加1
                updata(x1,y2+1,-1);//(x1,y2+1)~(n,n)矩阵整体减1
                updata(x2+1,y1,-1);//(x2+1,y1)~(n,n)矩阵整体减1
                //其中(x2+1,y2+1)~(n,n)矩阵被减去了两次
                updata(x2+1,y2+1,1);//将多减去的一次补回来
            }
            else if(s=='Q'){
                scanf("%d%d",&x1,&y1);
                printf("%d\n",getSum(x1,y1)%2);
            }
            getchar();
        }
        if(t>0) printf("\n");
    }
}