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.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner s = new Scanner(System.in);
		int N = s.nextInt();
		int K = s.nextInt();

		List<Pos> list = new ArrayList<>();
		for (int i = 0; i < N; i++) {
			int x = s.nextInt();
			int y = s.nextInt();
			// 白だったらどちらかにK移動すれば黒として扱える
			if ("W".equals(s.next())) {
				y += K;
			}
			// x,yでついの位置にあるものも同じ
			list.add(new Pos(x, y, K));
			list.add(new Pos(x + K, y + K, K));
		}
		
		int[][] space = new int[K * 2][K * 2];
		list.forEach(p -> ++space[p.x][p.y]);

		// y方向の和を求める
		int [][] spaceSum = new int[K * 2][K + 1];
		for(int x = 0; x < 2 * K; x++){
			int sum = 0;
			for(int y = 0; y < K; y++){
				sum += space[x][y];
			}
			spaceSum[x][0] = sum;
			for(int y = K; y < 2 * K; y++){
				sum = sum - space[x][y - K] + space[x][y];
				spaceSum[x][y - K + 1] = sum;
			}
		}
		
		// x方向の和を求め結果を出す
		int max = 0;
		for (int y = 0; y < K; y++) {
			int sum = 0;
			// はじめのk個
			for (int x = 0; x < K; x++) {
				sum += spaceSum[x][y];
			}
			if (sum > max) {
				max = sum;
			}
			for(int x = K; x < 2 * K; x++){
				sum = sum - spaceSum[x - K][y] + spaceSum[x][y];
				if (sum > max) {
					max = sum;
				}
			}
		}
		System.out.println(max);
	}

	public static class Pos {
		public final int x;
		public final int y;

		public Pos(int x, int y, int K) {
			this.x = x % (2 * K);
			this.y = y % (2 * K);
		}

		@Override
		public String toString() {
			return "Pos [x=" + x + ", y=" + y + "]";
		}
	}
}