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 Java7 (OpenJDK 1.7.0) (judge verdict: Wrong Answer)

import java.util.HashSet;
import java.util.Iterator;
import java.util.Scanner;

public class Main {
    private static int N,K;
    private static HashSet where;
    private static int[][][] mem;
    private static int[][][] cum;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        N = sc.nextInt();
        K = sc.nextInt();
        mem = new int[2*K][2*K][2];
        where = new HashSet();
        for (int i = 0; i < N; i++) {
            int x = sc.nextInt()%(2*K);
            int y = sc.nextInt()%(2*K);
            String c = sc.next();
            mem[y][x][c.equals("B") ? 1:0]++;
            where.add(new int[]{x,y});
        }


        cum = new int[2*K+1][2*K+1][2];
        Iterator<int[]> iter = where.iterator();
        while(iter.hasNext()) {
            int[] xy = iter.next();
            int x = xy[0];
            int y = xy[1];
            cum[y+1][x+1][0] = cum[y][x+1][0] + cum[y+1][x][0] - cum[y][x][0] + mem[y][x][0];
            cum[y+1][x+1][1] = cum[y][x+1][1] + cum[y+1][x][1] - cum[y][x][1] + mem[y][x][1];
        }

        int ans = 0;
        for(int y = 0; y < K; y++) {
            for(int x = 0; x < K; x++) {
                ans = Math.max(ans, eval(x,y));
            }
        }
        System.out.println(ans);
    }

    private static int count(int x1, int y1, int x2, int y2, int c) {
        return cum[y2][x2][c] - cum[y1][x2][c] - cum[y2][x1][c] + cum[y1][x1][c];
    }

    private static int eval(int x, int y) {
        int b1 = count(0,0,x,y,1) + count(K+x,0,2*K,y,1) + count(x,y,K+x,K+y,1) + count(0,K+y,x,2*K,1) + count(K+x,K+y,2*K,2*K,1);
        int w1 = count(x,0,K+x,y,0) + count(0,y,x,K+y,0) + count(K+x,y,2*K,K+y,0) + count(x,K+y,K+x,2*K,0);
        int b2 = count(0,0,x,y,0) + count(K+x,0,2*K,y,0) + count(x,y,K+x,K+y,0) + count(0,K+y,x,2*K,0) + count(K+x,K+y,2*K,2*K,0);
        int w2 = count(x,0,K+x,y,1) + count(0,y,x,K+y,1) + count(K+x,y,2*K,K+y,1) + count(x,K+y,K+x,2*K,1);
        return Math.max(b1+w1, b2+w2);
     }
}