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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {

    static int N,K;

    public static void main(String[] args){

        Scanner sc = new Scanner(System.in);

        N = sc.nextInt();
        K = sc.nextInt();

        int[] x = new int[N];
        int[] y = new int[N];
        char[] c = new char[N];

        for(int i=0;i<N;i++){
            x[i]=sc.nextInt();
            y[i]=sc.nextInt();
            c[i]=sc.next().charAt(0);
        }

        int max=0;
        int count1;
        int count2;
        for(int i=-K+1;i<K;i++){
            for(int j=-K+1;j<K;j++) {
                count1=1;
                count2=0;
                int bx=x[0]-i;
                int by=y[0]-j;
                for(int a=1;a<N;a++){
                    int rex=x[a]-bx;
                    int rey=y[a]-by;

                    int qx = (rex/K)%2;
                    int qy = (rey/K)%2;

                    if(qx==0 && qy==0){
                        if(c[0]==c[a])count1++;
                        else count2++;
                    } else if(qx==0 && qy==1){
                        if(c[0]!=c[a])count1++;
                        else count2++;
                    } else if(qx==1 && qy==0){
                        if(c[0]!=c[a])count1++;
                        else count2++;
                    } else if(qx==1 && qy==1){
                        if(c[0]==c[a])count1++;
                        else count2++;
                    }

                }
                max=Math.max(max,count1);
                max=Math.max(max,count2);
            }
        }

        System.out.println(max);

    }
}