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.*;


class CheckerBoard{
    private int patternSize;
    private int patternStartX, patternStartY;

    CheckerBoard(int s){
        this.patternSize = s;
        this.patternStartX = 0;
        this.patternStartY = 0;
    }

    public boolean checkPatternBoolean(int x, int y, boolean b){
        int xDiv = ((x + this.patternStartX) / this.patternSize) % 2;
        int yDiv = ((y + this.patternStartY) / this.patternSize) % 2;
        boolean isWhite = xDiv == yDiv;
        if(b == isWhite){
            return true;
        }else {
            return false;
        }
    }

    public void rotateX(int i){
        this.patternStartX += i;
    }
    public void rotateY(int i){
        this.patternStartY += i;
    }

    public int getPatternSize() {
        return patternSize;
    }

    public int getPatternStartX() {
        return patternStartX;
    }
    public int getPatternStartY() {
        return patternStartY;
    }
}


class Demand{
    int x;
    int y;
    boolean c;

    Demand(int x, int y, boolean c){
        this.x = x;
        this.y = y;
        this.c = c;
    }

    public int getX() {
        return x;
    }
    public int getY() {
        return y;
    }
    public boolean isC() {
        return c;
    }
}


public class Main {
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();

        CheckerBoard cb = new CheckerBoard(k);
        ArrayList<Demand> demands = new ArrayList<Demand>();
        for(int i = 0; i < n; i++){
            int x = sc.nextInt();
            int y = sc.nextInt();
            boolean c = sc.next().endsWith("W");
            demands.add(new Demand(x, y ,c));
        }

        int max = 0;

        for(int x = 0; x < cb.getPatternSize()+1; x++){
            for(int y = 0; y < cb.getPatternSize()+1; y++){
                int currentDemandCount = 0;
                for(Demand d : demands){
                    if(cb.checkPatternBoolean(d.getX(), d.getY(), d.isC())){

                        currentDemandCount++;
                    }
                }
                if(currentDemandCount > max){
                    max = currentDemandCount;
                }
                cb.rotateY(1);
            }
            cb.rotateY(-cb.getPatternSize());
            cb.rotateX(1);
        }

        System.out.println(max);
    }
}