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の座標に写像
                hope[i, 1] = int.Parse(tmphope[1]) % (2 * k);//2K*2Kの座標に写像
                if (tmphope[2] == "W") hope[i, 2] = 0; else { hope[i, 2] = 1; }//WBを01に変換
            }

            //希望数を列ごとに数える hopenum[yi]:yi列の希望数
            int[] hopenum = Enumerable.Repeat<int>(0, 2*k).ToArray();
            for (int p = 0; p < n; p++){ hopenum[hope[p, 1]]++; }

            //合致数チェック
            for (int xm = 0; xm < k; xm++)
            {
                int[] countline= Enumerable.Repeat<int>(0, 2*k).ToArray();// countline[yi]:yi列の合致数
                int sum =0;//合致数合計
                
                //ym==0のときの合致数を列ごとに数える
                for(int p=0; p<n; p++)
                {
                    if( Chk(hope[p,0],hope[p,1],xm,0,k)==hope[p,2]) 
                    {
                        countline[ hope[p,1] ]++;
                        sum++;
                    }


                    /*
                    Console.WriteLine("xi:"+ hope[p, 0]+" yi:"+ hope[p, 1]+" xm:"+ xm+ " ym:" + 0 + " k:" + k);
                    Console.WriteLine(Chk(hope[p, 0], hope[p, 1], xm, 0, k) == hope[p, 2]);

                    Console.Write("hopenum[ ");
                    for (int ym = 0; ym < k * 2; ym++)
                    {
                        Console.Write(hopenum[ym] + ", ");
                    }
                    Console.Write(" ]");
                    Console.Write("countline[ ");
                    for (int ym = 0; ym < k*2; ym++)
                    {
                        Console.Write(countline[ym]+", ");
                    }
                    Console.Write(" ]");
                    Console.WriteLine();
                    */

                }
                if (max < sum) { max = sum; }


                //列ごとに合致と非合致を反転させることで平行移動と同じ状態にする
                for (int ym =0; ym<k; ym++)
                {
                    sum = sum + hopenum[ym] - countline[ym] * 2;//ym列を反転
                    sum = sum + hopenum[ym+k] - countline[ym+k] * 2;//ym+k列を反転
                    if (max < sum) { max = sum; }
                }
                if(max >= n) { break; }
            }
            Console.WriteLine(max);
        }


        /* 座標(xi,yi)の色を返す　0,1の2色  
         * 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 */
        private static int Chk(int xi, int yi, int xm, int ym, int k)
        {
            if(xm <= xi && xi < k+xm && ym <= yi && yi < k + ym) { return 0; }
            if(k+xm <= xi && k+ym <= yi) { return 0; }
            if(k+xm <= xi && yi< ym) { return 0; }
            if(xi<xm && k+ym <= yi) { return 0; }
            return 1;
        }

    }

}