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: Runtime Error)

import java.util.Scanner;

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

        int max_x = 0;
        int max_y = 0;
        for (int i = 0; i < n; i++) {
            xs[i] = scanner.nextInt();
            if (xs[i] > max_x) {
                max_x = xs[i];
            }
            ys[i] = scanner.nextInt();
            if (ys[i] > max_y) {
                max_y = ys[i];
            }

            cs[i] = scanner.next().equals("B"); // black -> true, white -> false
        }

//        boolean[][] BWFlags = new boolean[max_x + k * 2][max_y + k];
        boolean[] x_flag = new boolean[max_x + k * 2];
        boolean[] y_flag = new boolean[max_x + k];
        for (int i_y = 0; i_y < max_y + k; i_y++) {
            for (int i_x = 0; i_x < max_x + k * 2; i_x++) {
                if ((i_y / k) % 2 == 0) {
                    if ((i_x / k) % 2 == 1) {
                        x_flag[i_x] = true;
                        y_flag[i_y] = true;
//                        BWFlags[i_x][i_y] = true;
                    }
                } else {
                    if ((i_x / k) % 2 == 0) {
                        x_flag[i_x] = true;
                        y_flag[i_y] = true;
//                        BWFlags[i_x][i_y] = true;
                    }
                }
            }
        }

        int max_able = 0;
        for (int k_x = 0; k_x < k * 2; k_x++) {
            for (int k_y = 0; k_y < k; k_y++) {

                int sum_able = 0;
                for (int i_n = 0; i_n < n; i_n++) {
                    int x = xs[i_n];
                    int y = ys[i_n];
                    boolean c = cs[i_n];

                    if (c == x_flag[x + k_x] && c == y_flag[y + k_y]) {
                        sum_able++;
                    }
//                    if (BWFlags[x + k_x][y + k_y] == c) {
//                        sum_able++;
//                    }
                }
                if (sum_able > max_able) {
                    max_able = sum_able;
                }
            }
        }

        System.out.print(max_able);
    }
}