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: Accepted)

using System;
using System.Linq;//リストの使用
using System.Collections.Generic;
using System.Text;//テキストの高速出力に必要
class Program
{
	static void Main()
	{
		string[] input = Console.ReadLine().Split(' ');
		int n = int.Parse(input[0]);
		int k = int.Parse(input[1]);
    long[][] vertexes = new long[n][];//ここに受け取る
    long[,] sumBlacks = new long[k,k];//0
    long[,] sumWhites = new long[k,k];//1
    long answer = 0;
    for(int i = 0; i < n; i++)
    {
      vertexes[i] = new long[3];
      var nums = Console.ReadLine().Split(' ');
      vertexes[i][0] = long.Parse(nums[0])%k;
      vertexes[i][1] = long.Parse(nums[1])%k;
      if(nums[2] == "B" && (long.Parse(nums[0])/k + long.Parse(nums[1])/k)%2 == 0 ||
        nums[2] == "W" && (long.Parse(nums[0])/k + long.Parse(nums[1])/k)%2 == 1)
      {
        vertexes[i][2] = 0;
      }else vertexes[i][2] = 1;
    }

    for(int i = 0; i < n; i++)
    {
      //Console.WriteLine(vertexes[i][0]+" "+vertexes[i][1]+" "+vertexes[i][2]+"aa");
      if(vertexes[i][2] == 0)
      {
        sumBlacks[vertexes[i][0],vertexes[i][1]]++;
      }else sumWhites[vertexes[i][0],vertexes[i][1]]++;
    }
    sumBlacks = Sum2D(sumBlacks);
    sumWhites = Sum2D(sumWhites);

    for(int i = 0; i <= k; i++)//(i,j)がBの右下
    {
      for(int j = 0; j <= k; j++)
      {
        long answerMemo = sumWhites[k,k];
        answerMemo -= sumWhites[i,j];
        answerMemo += sumBlacks[i,j];
        if(i <= k && j <= k)
        {
          answerMemo -= Sum2DAnswer(sumWhites,i,k-1,j,k-1);
          answerMemo += Sum2DAnswer(sumBlacks,i,k-1,j,k-1);
        }
        answer = Math.Max(answer,answerMemo);

        answerMemo = sumBlacks[k,k];
        answerMemo -= sumBlacks[i,j];
        answerMemo += sumWhites[i,j];
        if(i <= k && j <= k)
        {
          answerMemo -= Sum2DAnswer(sumBlacks,i,k-1,j,k-1);
          answerMemo += Sum2DAnswer(sumWhites,i,k-1,j,k-1);
        }
        answer = Math.Max(answer,answerMemo);

        //Console.WriteLine(i+" "+j+" "+answerMemo);
        
      }
    }

		Console.WriteLine(answer);
	}

  static long[,] Sum2D(long[,] nums)
  {//左上からi行j列の2次元累積和をO(hw)で返す
  //例えば{(1,2),(3,4)}なら{(0,0,0),(0,1,3),(0,4,10)}を返す
    int h = nums.GetLength(0);
    int w = nums.GetLength(1);
    long[,] answers = new long[h+1,w+1];

    for(int i = 1; i <= h; i++)
    {
      for(int j = 1; j <= w; j++)
      {
        answers[i,j] = answers[i,j-1]+answers[i-1,j]-answers[i-1,j-1]+nums[i-1,j-1];
      }
    }

    return answers;
  }

  static long Sum2DAnswer(long[,] sums, int hs, int hg, int ws, int wg)
  {//引数の2次元累積和に対し、0-indexで縦hs~ws横ws~wgの長方形内の和をO(1)で返す
    return (sums[hg+1,wg+1]-sums[hs,wg+1]-sums[hg+1,ws]+sums[hs,ws]);
  }
  
}