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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {

	public static void main(String[] args) {
		try (BufferedReader reader = new BufferedReader(
				new InputStreamReader(System.in))) {
			String[] input = reader.readLine().split(" ");
			int N = Integer.parseInt(input[0]);
			int K = Integer.parseInt(input[1]);
			int[] X = new int[N];
			int[] Y = new int[N];
			boolean[] C = new boolean[N];
			int maxX = 0;
			int maxY = 0;
			for (int i = 0; i < N; i++) {
				input = reader.readLine().split(" ");
				X[i] = Integer.parseInt(input[0]);
				maxX = maxX < X[i] ? X[i] : maxX;
				Y[i] = Integer.parseInt(input[1]);
				maxY = maxY < Y[i] ? Y[i] : maxY;
				C[i] = "W".equals(input[2]) ? false : true;
			}
			maxX++;
			maxY++;
			Boolean[][] expect = new Boolean[maxX][maxY];
			for (int i = 0; i < N; i++) {
				expect[X[i]][Y[i]] = C[i];
			}

			boolean[][] check = new boolean[maxX + K * 2][maxY + K * 2];
			for (int i = 0; i < maxX + K * 2; i++) {
				for (int j = 0; j < maxY + K * 2; j++) {
					check[i][j] = i % (2 * K) < K && j % (2 * K) < K
							|| i % (2 * K) >= K && j % (2 * K) >= K;
				}
			}

			int max = 0;
			for (int i = 0; i < K * 2; i++) {
				for (int j = 0; j < K * 2; j++) {
					int count = count(check, i, j, expect);
					max = max < count ? count : max;
				}
			}
			System.out.println(max);
		}
		catch (IOException e) {
			e.printStackTrace();
		}
	}

	private static int count(boolean[][] check, int stX, int stY, Boolean[][] expect) {
		int count = 0;
		for (int i = 0; i < expect.length; i++) {
			for (int j = 0; j < expect[i].length; j++) {
				if (expect[i][j] != null && check[i + stX][j + stY] == expect[i][j]) {
					count++;
				}
			}

		}
		return count;
	}
}