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 Java (OpenJDK 11.0.6) (judge verdict: Time Limit Exceeded)


import java.util.*;

public class Main {

	public static void main(String[] args) {
		Main main = new Main();
		main.run();
	}

	public void run() {
		Scanner sc = new Scanner(System.in);
		//2k 0～k-1=w, k～2k-1=b
		int n=sc.nextInt();
		int K=sc.nextInt();

		long x[] = new long[n];
		long y[] = new long[n];
		int c[] = new int[n];
		for(int i=0; i<n; i++) {
			x[i]=sc.nextLong();
			y[i]=sc.nextLong();
			if(sc.next().charAt(0)=='W') {
				c[i]=0;
			}else {
				c[i]=1;
			}
		}

		//W=0, B=1
		long ans = 0;
		for(int i=0; i<2*K; i++) {
			for(int j=0; j<2*K; j++) {
				int cc=0;
				int cnt=0;
				for(int k=0; k<n; k++) {
					if((x[k]+i)%(2*K)<K) {
						cc=0;
					}else {
						cc=1;
					}
					if((y[k]+j)%(2*K)>=K) {
						cc = (cc+1)%2;
					}
					if(cc==c[k]) {
						cnt++;
					}
				}
				ans = Math.max(ans, cnt);
			}
		}
		System.out.println(ans);
		sc.close();
	}
}