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.*;
import java.io.*;
public class Main {

    public static void main(String[] args) {
        Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        String[] s = in.nextLine().split(" ");
        int k = new Integer(s[1]);
        int n = new Integer(s[0]);
        int[][] point = new int[n][3];
        for (int num = 0; num < n; num++) {
            String[] d = in.nextLine().split(" ");
            point[num][0] = new Integer(d[0]);
            point[num][1] = new Integer(d[1]);
            point[num][2] = (d[2].equals("B")) ? 1 : 0;

        }

        int maxS = 0, minx, xx, yy, count, miny;
        for (int i = 0; i < k; i++) {
            minx = k;
            for (int j = 0; j < k; j++) {
                count = 0;
                miny = k;
                for (int[] p : point) {
                    xx = p[0] - i + k;
                    yy = p[1] - j + k;
                    minx = Math.min(xx % k, Math.min(minx, k - (xx % k) - 1));
                    miny = Math.min(yy % k, Math.min(miny, k - (yy % k) - 1));
                    if (((xx / k + yy / k) & 1) == p[2]) {
                        count++;
                    }
                }
                maxS = Math.max(maxS, Math.max(count, n - count));
                if (maxS == n) {
                    System.out.println(n);
                    return;
                }
                j += miny;
            }
            i += minx;
        }
        System.out.println(maxS);
    }
}