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.Collections;
using System.Collections.Generic;
using System.Linq;
using System.IO;
using System.Text;
using System.Globalization;
using System.Diagnostics;
using System.Threading.Tasks;



class Myon
{
    public Myon() { }
    public static int Main()
    {
        new Myon().calc();
        return 0;
    }
    

    Scanner cin;

    void calc()
    {
        cin = new Scanner();
        int N = cin.nextInt();
        int K = cin.nextInt();
        int[] x = new int[N];
        int[] y = new int[N];
        int[] y2 = new int[N];
        int[] y3 = new int[N];
        int[] z = new int[N];
        for (int i = 0; i < N; i++)
        {
            x[i] = cin.nextInt() % (2 * K) + (2 * K) % (2 * K);
            y[i] = cin.nextInt() % (2 * K) + (2 * K) % (2 * K);
            y2[i] = y[i] / K;
            y3[i] = y[i] % K;
            z[i] = cin.next() == "B" ? 0 : 1;
        }

        int bestAns = 0;
        for (int X = 0; X < K; X++)
        {
            int[] now = new int[N];
            int tmp = 0;
            int[] Add = new int[2 * K];

            for (int i = 0; i < N; i++)
            {
                int first = ((x[i] + X) / K & 1) ^ z[i] ^ (y2[i]);
                int Y = y3[i];
                if(first == 1)
                {
                    Add[Y]--;
                    Add[K + Y]++;
                }
                else
                {
                    Add[Y]++;
                    Add[K + Y]--;
                }
                tmp += first;
            }
            int ans = tmp;
            for (int i = 0; i < 2 * K; i++)
            {
                tmp += Add[i];
                ans = Math.Max(ans, tmp);
            }


            bestAns = Math.Max(bestAns, ans);
        }

        Console.WriteLine(bestAns);
    }
}




class Scanner
{
    string[] s;
    int i;

    char[] cs = new char[] { ' ' };

    public Scanner()
    {
        s = new string[0];
        i = 0;
    }

    public string next()
    {
        if (i < s.Length) return s[i++];
        string st = Console.ReadLine();
        while (st == "") st = Console.ReadLine();
        s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
        if (s.Length == 0) return next();
        i = 0;
        return s[i++];
    }

    public int nextInt()
    {
        return int.Parse(next());
    }
    public int[] ArrayInt(int N, int add = 0)
    {
        int[] Array = new int[N];
        for (int i = 0; i < N; i++)
        {
            Array[i] = nextInt() + add;
        }
        return Array;
    }

    public long nextLong()
    {
        return long.Parse(next());
    }

    public long[] ArrayLong(int N, long add = 0)
    {
        long[] Array = new long[N];
        for (int i = 0; i < N; i++)
        {
            Array[i] = nextLong() + add;
        }
        return Array;
    }

    public double nextDouble()
    {
        return double.Parse(next());
    }


    public double[] ArrayDouble(int N, double add = 0)
    {
        double[] Array = new double[N];
        for (int i = 0; i < N; i++)
        {
            Array[i] = nextDouble() + add;
        }
        return Array;
    }
}