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;
import java.util.Collections;

class Main{
    public static void main(String[] args){
        Scanner scan = new Scanner(System.in);

        int N = scan.nextInt();
        int K = scan.nextInt();
        int[][] map = new int[4*K][4*K];
        for(int i=0;i<N;++i){
            int x = scan.nextInt();
            int y = scan.nextInt();
            if(scan.next().equals("W"))x+=K;
            y%=2*K;x%=2*K;
            map[y][x]++;
        }
        for(int i=0;i<2*K;++i)for(int j=0;j<2*K;++j){
            map[i][j+2*K]=map[i][j];
            map[i+2*K][j]=map[i][j];
            map[i+2*K][j+2*K]=map[i][j];
        }

        int[][] sum = new int[4*K+1][4*K+1];
        for(int i=0;i<4*K;++i)for(int j=0;j<4*K;++j){
            sum[i+1][j+1]=sum[i][j+1]+sum[i+1][j]-sum[i][j]+map[i][j];
        }

        int ans = 0;
        for(int i=1;i<=K;++i)for(int j=1;j<=K;++j){
            ans = Math.max(ans, sum[i+2*K-1][j+2*K-1]-sum[i+K-1][j+2*K-1]-sum[i+2*K-1][j+K-1]+sum[i+K-1][j+K-1]
                               +sum[i+K-1][j+K-1]    -sum[i-1][j+K-1]    -sum[i+K-1][j-1]    +sum[i-1][j-1]);
        }
        System.out.println(ans);


    }

}