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>

int main()
{
    int n,k;
    scanf("%d %d", &n, &k);
    int x[n],y[n];
    char c[n];
    for(int i=0;i<n;i++){
        scanf("%d %d %c", &x[i], &y[i], &c[i]);
    }
    int res=0,sres,dres;
    int color=0, p,black;
    // Black
    //for(color=0;color<2;color++){
        for(int i = 0; i < k; i++){
            for(int j = 0; j < k; j++){
                sres=0;dres=0;
                for(int l = 0; l < n; l++){
                    p=(((x[l]+i)/k)+((y[l]+j)/k));
                    black=(p%2==1);
                    if(c[l]=='B'){
                        if(black){
                            sres+=1;
                        }
                    }else{
                        if(!black){
                            sres+=1;
                        }
                    }
                    if(c[l]=='B'){
                        if(!black){
                            dres+=1;
                        }
                    }else{
                        if(black){
                            dres+=1;
                        }
                    }
                }
                if(res<sres){
                    res = sres;
                }
                if(res<dres){
                    res=dres;
                }
                if(res==n){
                    break;
                }
            }
        }
    //}
    printf("%d\n", res);
    return 0;
}