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;

public class Main {
	public static void main(String[] args) {
		final Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int k = sc.nextInt();
		long[] x = new long[n], y = new long[n];
		boolean[] b = new boolean[n];//Wならtrue、Bならfalse
		for (int i = 0; i < n; i++) {
			x[i] = sc.nextLong() % (2 * k);
			y[i] = sc.nextLong() % (2 * k);
			if (sc.next().equals("W")) {
				//nothing do
			} else {
				x[i] -= k;
				y[i] -= k;
				if (x[i] < 0) {
					x[i] += (2 * k);
				}
				if (y[i] < 0) {
					y[i] += (2 * k);
				}
			}
		}
		long ans = 0L;
		for (int i = 0; i <= k+1; i++) {
			for (int m = 0; m <= k+1; m++) {
				long count = 0L;
				for (int j = 0; j < n; j++) {
					if (x[j] >= i && x[j] <= i + k && y[j] >= m && y[j] <= m + k) {
						count++;
					}
				}
				ans = Math.max(ans, count);
			}
		}
		System.out.println(ans);
	}
}