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 {

  private static final int Black = 0;
  private static final int White = 1;

  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];
    int[] y = new int[N];
    int[] c = new int[N];

    final String BlackColor = "B";
    final int K2 = 2 * K;
    for (int i = 0; i < N; i++) {
      x[i] = scanner.nextInt() % K2;
      y[i] = scanner.nextInt() % K2;
      c[i] = BlackColor.equals(scanner.next()) ? Black : White;
    }
    new Main().execute(N, K, x, y, c);
  }

  public void execute(int N, int K, int[] x, int[] y, int[] c) {
    final int K2 = 2 * K;
    int maxDesires = 0;
    for (int i = 0; i < K2; i++) {
      for (int j = 0; j < K2; j++) {
        // Assumes cell (i,j) is black and bottom-left corner of (K,K) square
        int count = 0;
        for (int n = 0; n < N; n++) {
          if (c[n] == getColor(K, i, j, x[n], y[n])) {
            count++;
          }
        }

        if (maxDesires < count) {
          maxDesires = count;
        }
      }
    }

    System.out.println(maxDesires);
  }

  /**
   * Assume bottom-left black corner is (rootX, rootY), get color of cell (x,y)
   */
  public int getColor(int K, int rootX, int rootY, int x, int y) {
    int dx = x - rootX;
    int dy = y - rootY;
    if (dx < 0) {
      dx -= K - 1;
    }
    if (dy < 0) {
      dy -= K - 1;
    }

    return ((dx / K + dy / K) & 1) == 1 ? Black : White;
  }

}