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;

class Solver {
  public void solve() {
    var buf = Console.ReadLine().Split();
    int n = int.Parse(buf[0]);
    int k = int.Parse(buf[1]);
    int m = k * 2;
    int[,,] a = new int[2, m + 1, m + 1];
    for (int i = 0; i < n; i++) {
      buf = Console.ReadLine().Split();
      int x = int.Parse(buf[0]);
      int y = int.Parse(buf[1]);
      int c = buf[2]=="B" ? 1 : 0;
      x %= m; y %= m;
      a[c,x,y]++;
    }
    for (int c = 0; c < 2; c++) {
      for (int i = 0; i < m; i++) {
        for (int j = 0; j <= m; j++) {
          a[c, i + 1, j] += a[c, i, j];
        }
      }
      for (int i = 0; i <= m; i++) {
        for (int j = 0; j < m; j++) {
          a[c, i, j + 1] += a[c, i, j];
        }
      }
    }
    int ans = 0;
    for (int i = 0; i < k; i++) {
      for (int j = 0; j < k; j++) {
        for (int c = 0; c < 2; c++) {
          int cur = 0;
          int ti = i + k, tj = j + k;
          for (int nc = 0; nc < 2; nc++) {
            int tmp = 0;
            tmp += a[nc, m, tj] - a[nc, m, j];
            tmp += a[nc, ti, m] - a[nc, i, m];
            tmp -= (a[nc, ti, tj] - a[nc, ti, j] - a[nc, i, tj] + a[nc, i, j]) * 2;
            if (c != nc) tmp = a[nc, m, m] - tmp;
            cur += tmp;
          }
          ans = Math.Max(ans, cur);
        }
      }
    }
    Console.WriteLine(ans);
  }
}

/* Main */
class ReSha {
  static void Main() {
    new Solver().solve();
  }
}