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) {
    Scanner scanner = new Scanner(System.in);

    int N = scanner.nextInt();
    int K= scanner.nextInt();

    int count = 0;
    int[][] delta = new int[K][K];
    for (int i = 0; i < N; i++) {
      int x = scanner.nextInt();
      int y = scanner.nextInt();
      char c = scanner.next().charAt(0);
      int p = x / K + y / K;

      delta[x % K][y % K]++;
      if (p % 2 == 0 && c == 'W' || p % 2 == 1 && c == 'B') {
        delta[x % K][y % K] -= 2;
        count++;
      }
    }

    for (int i = 0; i < K; i++) for (int j = 1; j < K; j++) delta[i][j] += delta[i][j - 1];
    for (int j = 0; j < K; j++) for (int i = 1; i < K; i++) delta[i][j] += delta[i - 1][j];

    int max = count;
    for (int i = 0; i <= K; i++) {
      for (int j = 0; j <= K; j++) {
        int c = count;
        if (i > 0) c += delta[i - 1][K - 1];
        if (j > 0) c += delta[K - 1][j - 1];
        if (i > 0 && j > 0) c -= 2 * delta[i - 1][j - 1];
        max = Math.max(max, c);
      }
    }

    System.out.println(max);
  }
}