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: Time Limit Exceeded)

import java.io.*;
import java.util.*;

class Main {

	public static void main(String[] args) throws Exception {
		final Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int K = sc.nextInt();
		int[] x = new int[N];
		int[] y = new int[N];

		String[] c = new String[N];
		for (int i = 0; i < N; i++) {
			x[i] = sc.nextInt() % (K * 2);
			y[i] = sc.nextInt() % (K * 2);
			c[i] = sc.next();
		}

		int answer = 0;
		for (int i = 0; i < K; i++) {
			for (int j = 0; j < K; j++) {
				int ab = 0, aw = 0;
				int bb = 0, bw = 0;
				for (int k = 0; k < N; k++) {
					if ((x[k] + i) % (2 * K) >= K) {
						if ((y[k] + j) % (2 * K) >= K) {
							// b
							if (c[k].charAt(0) == 'B') {
								bb++;
							} else {
								bw++;
							}
						} else {
							// a
							if (c[k].charAt(0) == 'B') {
								ab++;
							} else {
								aw++;
							}
						}
					} else {
						if ((y[k] + j) % (2 * K) >= K) {
							// a
							if (c[k].charAt(0) == 'B') {
								ab++;
							} else {
								aw++;
							}
						} else {
							// b
							if (c[k].charAt(0) == 'B') {
								bb++;
							} else {
								bw++;
							}
						}
					}

					answer = Math.max(answer, aw + bb);
					answer = Math.max(answer, ab + bw);
				}
			}
		}

		System.out.println(answer);
	}
}