Problem:
Score : 500 points

Problem Statement

AtCoDeer is thinking of painting an infinite two-dimensional grid in a checked pattern of side K.
Here, a checked pattern of side K is a pattern where each square is painted black or white so that each connected component of each color is a K × K square.
Below is an example of a checked pattern of side 3:

AtCoDeer has N desires.
The i-th desire is represented by x_i, y_i and c_i.
If c_i is B, it means that he wants to paint the square (x_i,y_i) black; if c_i is W, he wants to paint the square (x_i,y_i) white.
At most how many desires can he satisfy at the same time?

Constraints

1 ≤ N ≤ 10^5

1 ≤ K ≤ 1000

0 ≤ x_i ≤ 10^9

0 ≤ y_i ≤ 10^9

If i ≠ j, then (x_i,y_i) ≠ (x_j,y_j).

c_i is B or W.

N, K, x_i and y_i are integers.

Input

Input is given from Standard Input in the following format:

N K
x_1 y_1 c_1
x_2 y_2 c_2
:
x_N y_N c_N

Output

Print the maximum number of desires that can be satisfied at the same time.

Sample Input 1

4 3
0 1 W
1 2 W
5 3 B
5 4 B

Sample Output 1

4

He can satisfy all his desires by painting as shown in the example above.

Sample Input 2

2 1000
0 0 B
0 1 W

Sample Output 2

2

Sample Input 3

6 2
1 2 B
2 1 W
2 2 B
1 0 B
0 6 W
4 5 W

Sample Output 3

4

Solution in C (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<stdio.h>
#include<math.h>
int main(void){
    int x[100000];
    int y[100000];
    char c[100000];

    int N,K;
    
    int ab=0;
    int aw=0;
    int ans=0;
    
    
    scanf("%d %d",&N,&K);
    int i,j,l;
    
    for(i=0;i<N;i++){
        scanf("%d %d %s",&x[i],&y[i],&c[i]);
    }
    for(i=0;i<K;i++){
        for(j=0;j<K;j++){
            for(l=0;l<N;l++){
                if((2*K+x[l]-i)%(2*K)<K){
                    if((2*K+y[l]-j)%(2*K)<K){
                        if(c[l]=='B'){
                            ab++;
                        }else{
                            aw++;
                        }
                    }else{
                        if(c[l]=='W'){
                            ab++;
                        }else{
                            aw++;
                        }
                    }
                
                }else{
                    if((2*K+y[l]-j)%(2*K)<K){
                        if(c[l]=='W'){
                            ab++;
                        }else{
                            aw++;
                        }
                    }else{
                        if(c[l]=='B'){
                            ab++;
                        }else{
                            aw++;
                        }    
                    }
                }                
            }
            if(ab<aw){
                ab=aw;
            }
            if(ans<ab){
                ans = ab;
                if(ans==N){
                        printf("%d",ans);
                    return 0;
                    }
                    
            }
            ab=0;aw=0;
        }
    }
    printf("%d",ans);
    return 0;
}