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: Accepted)

object Main extends App {
  val Array(n, k) = io.StdIn.readLine().split(' ').map(_.toInt)
  val k2 = k*2
  val end = k2-1
  // 0 ≦ x,y ≦ 2K に圧縮
  val squares = Array.fill(k2)(Array.fill(k2)(0))
  for (i <- 1 to n) {
    val Array(x, y, color) = io.StdIn.readLine().split(' ')
    // 'W'はY方向にKずらした'B'と同値
    squares(x.toInt % k2)((y.toInt + (if (color == "B") 0 else k)) % k2) += 1
  }
  // 計算量削減のために累積和を事前算出
  for {
    i <- 0 to end
    j <- 1 to end
  } squares(i)(j) += squares(i)(j-1)
  for {
    j <- 0 to end
    i <- 1 to end
  } squares(i)(j) += squares(i-1)(j)
  // ex) k = 2    
  //      □ □ ■ ■
  //    k □ □ ■ ■
  //      ■ ■ □ □
  //    0 ■ ■ □ □
  //      0   k 
  val cnt =
    (squares(k-1)(k-1)) +
    (squares(end)(end) - squares(k-1)(end) - squares(end)(k-1) + squares(k-1)(k-1))
  var max = Math.max(cnt, n - cnt)
  for {
    i <- 0 to k-1
    j <- 0 to k-1
  } {
    // ex) k = 2
    //      ■ □ □ ■ 
    //  j+k □ ■ ■ □
    //      □ ■ ■ □
    //    j ■ □ □ ■
    //      i   i+k
    val cnt =
      (squares(i)(j)) +
      (squares(end)(j) - squares(i+k)(j)) +
      (squares(i+k)(j+k) - squares(i+k)(j) - squares(i)(j+k) + squares(i)(j)) +
      (squares(i)(end) - squares(i)(j+k)) +
      (squares(end)(end) - squares(end)(j+k) - squares(i+k)(end) + squares(i+k)(j+k))
    max = Math.max(Math.max(cnt, n - cnt), max)
  }
  print(max)
}