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.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);
		int N = scanner.nextInt();
		int K = scanner.nextInt();
		int x[] = new int[N+1];
		int y[] = new int[N+1];
		boolean c[] = new boolean[N+1];
		for (int i = 1; i <= N; i++) {
			x[i] = scanner.nextInt();
			y[i] = scanner.nextInt();
			c[i] = scanner.next().equals("B");
		}
		scanner.close();
		
		int answer = 0;
		int answer2 = 0;
		int FA = 0;
		
		for (int heightSlide = 0; heightSlide < K; heightSlide++) {
			for (int widthSlide = 0; widthSlide < K; widthSlide++) {
				answer = 0;
				answer2 = 0;
				for (int i = 1; i <= N; i++) {
					if (((((x[i] + heightSlide) % (K*2)) - K < 0) ^ (((y[i] + widthSlide) % (K*2)) - K < 0)) == c[i]) {
						answer++;
					} else {
						answer2++;
					}
				}
				FA = Math.max(FA, answer);
				FA = Math.max(FA, answer2);
			}
		}
		
		System.out.println(FA);
		
	}

}