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.Collections.Generic;
using System.Linq;
using System.IO;
using System.Text;
using System.Text.RegularExpressions;
using System.Globalization;
using System.Diagnostics;
using static System.Console;
//using System.Numerics;
//using static System.Math;

class Program
{
    static void Main()
    {
        //SetOut(new StreamWriter(OpenStandardOutput()) { AutoFlush = false });
        new Program().solve();
        Out.Flush();
    }
    Scanner cin = new Scanner();
    readonly int[] dd = { 0, 1, 0, -1, 0 }; //→↓←↑
    readonly int mod = 1000000007;

    int N, K;
    void solve()
    {
        N = cin.nextint;
        K = cin.nextint;
        var A = new int[2 * K, 2 * K];
        var B = new int[2 * K, 2 * K];
        for (int i = 0; i < N; i++)
        {
            int y = cin.nextint % (2 * K);
            int x = cin.nextint % (2 * K);
            var c = cin.next;
            if (c == "W")
            {
                A[y, x]++;
            }
            else
            {
                B[y, x]++;
            }
        }

        var C = new int[2 * K, 2 * K];
        var D = new int[2 * K, 2 * K];
        for (int i = 0; i < 2 * K; i++)
        {
            for (int j = 0; j < 2 * K; j++)
            {
                int y = (i + K) % (2 * K);
                int x = (j + K) % (2 * K);
                C[i, j] = A[i, j] + A[y, x] + B[y, j] + B[i, x];
                D[i, j] = B[i, j] + B[y, x] + A[y, j] + A[i, x];
            }
        }

        var E = new int[2 * K, 2 * K];
       //WriteLine("C");
        for (int i = 0; i < 2 * K; i++)
        {
            for (int j = 0; j < 2 * K; j++)
            {
                //Write(C[i, j]);
                E[i, j] = C[i, j];
            }
           //WriteLine();
        }


      
        for (int i = 0; i < 2 * K; i++)
        {
            for (int j = 0; j < 2 * K; j++)
            {
                if (j == 0)
                {

                }
                else
                {
                    E[i, j] += E[i, j - 1];
                    //if (j >= K) E[i, j] -= E[i, j - K];
                }
            }
        }


        for (int i = 0; i < 2 * K; i++)
        {
            for (int j = 0; j < 2 * K; j++)
            {
                if (i == 0)
                {

                }
                else
                {
                    E[i, j] += E[i - 1, j];
                   // if (i >= K) E[i, j] -= E[i - K, j];
                }
            }
        }
       // WriteLine("C");


        int ans = 0;
        for (int i = 0; i < 2 * K; i++)
        {
            for (int j = 0; j < 2 * K; j++)
            {
                //Write(E[i, j]);
                int d = E[i, j];
                if (i >= K) d -= E[i - K, j];
                if (j >= K) d -= E[i, j - K];
                if (i >= K && j >= K) d += E[i - K, j - K];
                ans = Math.Max(ans, d);
                //Write(d);
            }
            //WriteLine();
        }
        WriteLine(ans);

    }

}

class Scanner
{
    string[] s; int i;
    char[] cs = new char[] { ' ' };
    public Scanner() { s = new string[0]; i = 0; }
    public string[] scan { get { return ReadLine().Split(); } }
    public int[] scanint { get { return Array.ConvertAll(scan, int.Parse); } }
    public long[] scanlong { get { return Array.ConvertAll(scan, long.Parse); } }
    public double[] scandouble { get { return Array.ConvertAll(scan, double.Parse); } }
    public string next
    {
        get
        {
            if (i < s.Length) return s[i++];
            string st = ReadLine();
            while (st == "") st = ReadLine();
            s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
            i = 0;
            return next;
        }
    }
    public int nextint { get { return int.Parse(next); } }
    public long nextlong { get { return long.Parse(next); } }
    public double nextdouble { get { return double.Parse(next); } }
}