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;//リストの使用
using System.Collections.Generic;
class Program
{
	static void Main()
	{
		string[] input = Console.ReadLine().Split(' ');//Splitで区切り文字を指定して複数個受け取る。
		long n = long.Parse(input[0]);
		long k = long.Parse(input[1]);
    long[,] vertexes = new long[n,2];
    long[,] masses = new long[2*k,2*k];
    long answer = 0;
    for(long i = 0; i < n; i++)
    {
      var nums = Console.ReadLine().Split(' ');//１行ごとに受け取る
      vertexes[i,0] = long.Parse(nums[0]);
      vertexes[i,1] = long.Parse(nums[1]);
      vertexes[i,0] %= 2 * k;
      vertexes[i,1] %= 2 * k;
      if(nums[2] == "W")
      {
        if(vertexes[i,0] < k) vertexes[i,0] += k;//WをBに反転
        else vertexes[i,0] -= k;
        if(vertexes[i,1] < k) vertexes[i,1] += k;
        else vertexes[i,1] -= k;
      }
      /*
        Console.WriteLine(masses[0,0]+" "+masses[0,1]+" "+masses[0,2]+" "+masses[0,3]
        +" "+masses[1,0]+" "+masses[1,1]+" "+masses[1,2]+" "+masses[1,3]
        +" "+masses[2,0]+" "+masses[2,1]+" "+masses[2,2]+" "+masses[2,3]
        +" "+masses[3,0]+" "+masses[3,1]+" "+masses[3,2]+" "+masses[3,3]);
      */
      masses[vertexes[i,0], vertexes[i,1]]++;
      if(vertexes[i,0]>0 && vertexes[i,1]>0)//左下
      {
        masses[0, 0]++;
        masses[vertexes[i,0], vertexes[i,1]]++;
        masses[0, vertexes[i,1]]--;
        masses[vertexes[i,0], 0]--;
      }
      if(vertexes[i,0]>0 && vertexes[i,1]<k)//左上
      {
        masses[0, vertexes[i,1]+k]++;
        masses[vertexes[i,0], vertexes[i,1]+k]--;
      }
      if(vertexes[i,0]<k && vertexes[i,1]<k)//右上
      {
        masses[vertexes[i,0]+k, vertexes[i,1]+k]++;
      }
      if(vertexes[i,0]<k && vertexes[i,1]>0)//右下
      {
        masses[vertexes[i,0]+k, vertexes[i,1]]--;
        masses[vertexes[i,0]+k, 0]++;
      }
    }

    for(long i = 1; i < 2*k; i++)//右側へたす
    {
      for(long j = 0; j < 2*k; j++)
      {
        masses[i,j] += masses[i-1,j];
      }
    }

    for(long j = 1; j < 2*k; j++)//上側へたす
    {
      for(long i = 0; i < 2*k; i++)
      {
        masses[i,j] += masses[i,j-1];
      }
    }

    for(long i = 0; i < 2*k; i++)
    {
      for(long j = 0; j < 2*k; j++)
      {
        answer = Math.Max(answer, masses[i,j]);
      }
    }
    
		Console.WriteLine(answer);
	}
}