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.Scanner;

public class Main {

	public static void main(String[] args) {
		@SuppressWarnings("resource")
		Scanner stdin = new Scanner(System.in);

		int inN = stdin.nextInt();
		int inK = stdin.nextInt();

		int[] gridX = new int[inN];
		int[] gridY = new int[inN];
		for (int i = 0; i < inN; i++) {
			long inX = stdin.nextLong();
			long inY = stdin.nextLong();
			String inC = stdin.next();
			if (inC.equals("W")) {
				inX += inK;
			}

			gridX[i] = (int)(inX % (inK * 2));
			gridY[i] = (int)(inY % (inK * 2));
		}

		int max = 0;
		for (int blackX = 0; blackX < (inK * 2); blackX++) {
			for (int blackY = 0; blackY < (inK * 2); blackY++) {
				int count = 0;
				for (int i = 0; i < inN; i++) {
					if (isBlack(gridX[i], gridY[i], blackX, blackY, inK)) {
						count++;
					}
				}

				max = Math.max(max, count);
			}
		}

		System.out.println(max);
	}

	private static boolean isBlack(int x, int y, int blackX, int blackY, int k) {
		if (x < blackX) {
			x += (k * 2);
		}
		if ((x < blackX) || ((blackX + k) <= x)) {
			return false;
		}

		if (y < blackY) {
			y += (k * 2);
		}
		if ((y < blackY) || ((blackY + k) <= y)) {
			return false;
		}

		return true;
	}
}