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 C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Linq;

namespace D_Checker
{
    class Program
    {
        static void Main(string[] args)
        {
            String[] nktmp = Console.ReadLine().Split(' ');
            int n = int.Parse(nktmp[0]);
            int k = int.Parse(nktmp[1]);
            int[,] hope = new int[n, 3];
            int max = 0;


            //希望の読み込み
            for (int i = 0; i < n; i++)
            {
                string[] tmphope = Console.ReadLine().Split(' ');
                hope[i, 0] = int.Parse(tmphope[0]) % (2*k);//2K*2Kの座標に転写
                hope[i, 1] = int.Parse(tmphope[1]) % (2 * k);//2K*2Kの座標に転写
                if (tmphope[2] == "W") hope[i, 2] = 0; else { hope[i, 2] = 1; }//WBを01に変換
            }


            //希望合致数チェック
            for (int xm = 0; xm < k; xm++)
            {
                for(int ym =0; ym<k; ym++)
                {
                    int count = 0;
                    for(int p=0; p<n; p++)
                    {
                        if (chk(hope[p, 0], hope[p, 1], xm, ym, k) == hope[p, 2]){ count++; }
                    }
                    if(n - count > count) { count = n - count; }//色反転と比較して多い方を採用
                    if(count > max) { max = count; }//最大なら更新
                    if (max >= n) { break; }//打ち切り                        
                }
                if (max >= n) { break; }//打ち切り
            }

            Console.WriteLine(max);

        }


        /* 座標(xi,yi)の色を返す　白なら0 黒なら1  
         * xm,ym:市松模様が0,0からどれくらいずれているか  k:市松模様の一辺 */
        static int chk(int xi, int yi, int xm, int ym, int k)
        {
            if (  k+xm < xi && k+ym < yi) { return 0; }
            if (xm <= xi && xi < k + xm
                && ym <= yi && yi < k + ym) { return 0; }
            return 1;
        }

    }




}