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 Scala (2.11.7) (judge verdict: Wrong Answer)

object Main extends App {
  val Array(n, k) = io.StdIn.readLine().split(' ').map(_.toInt)
  val k2 = k * 2
  var (sum, white) = (0, 0)
  val matrix = new Array[Array[Int]](k2)
  (0 to k2-1).foreach(i => matrix(i) = new Array[Int](k2))
  (0 to n-1).foreach { i =>
    val temp = io.StdIn.readLine().split(' ')
      if (temp(2) == "B") {
        sum = sum + 1
        matrix(temp(0).toInt % k2)(temp(1).toInt % k2) = 1
      }
      else {
        sum = sum - 1
        white = white + 1
        matrix(temp(0).toInt % k2)(temp(1).toInt % k2) = -1
      }
  }
  for {
    i <- (0 to k2-1)
    j <- (1 to k2-1)
  } {
    val back = matrix(i)(j-1)
    matrix(i)(j) = matrix(i)(j) + back
  }
  for {
    i <- (1 to k2-1)
    j <- (0 to k2-1)
  } {
    val back = matrix(i-1)(j)
    matrix(i)(j) = matrix(i)(j) + back
  }
  val cnt =
      (matrix(k-1)(k-1)) +
      (matrix(k2-1)(k2-1) - matrix(k-1)(k2-1) - matrix(k2-1)(k-1) + matrix(k-1)(k-1))
  var max = Math.max(cnt, sum - cnt)
  for { 
    i <- (0 to k-1)
    j <- (0 to k-1)
  } {
    val cnt = 
      (matrix(i)(j)) +
      (matrix(i+k)(j+k) - matrix(i+k)(j) - matrix(i)(j+k) + matrix(i)(j)) +
      (matrix(k2-1)(j) - matrix(i+k)(j)) +
      (matrix(i)(k2-1) - matrix(i)(j+k)) +
      (matrix(k2-1)(k2-1) - matrix(k2-1)(j+k) - matrix(i+k)(k2-1) + matrix(i+k)(j+k))
    max = Math.max(Math.max(cnt, sum - cnt), max)
  }
  print(max + white)
}