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 Ruby (2.3.3) (judge verdict: Time Limit Exceeded)

n, k = gets.chomp.split(" ").map(&:to_i)
hopes = []
n.times {
  hopes.push (gets.chomp.split(" "))
}

hopes.each {|hope|
}

first = TRUE
max = 0
for i in 0..(k-1) do
  for j in 0..(k-1) do
    count = 0
    for hope in hopes do
      if first then
        hope[0] = hope[0].to_i % (2 * k)
        if hope[0] >= k then
          hope[0] = hope[0] - k
          hope[2] = if hope[2] == "W" then "B" else "W" end
        end
        hope[1] = hope[1].to_i % (2 * k)
        if hope[1] >= k then
          hope[1] = hope[1] - k
          hope[2] = if hope[2] == "W" then "B" else "W" end
        end
        hope[2] = hope[2] == "W"
      end
      parity = (hope[0] <= i) == (hope[1] <= j)
      if hope[2] == parity then
        count = count + 1
      end
    end
    first = FALSE
    if count > max then
      max = count
    end
    if (n - count) > max then
      max = n - count
    end
    if max == n then
      puts n
      exit
    end
  end
end
puts max