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: Wrong Answer)

N, K = gets.chomp.split.map(&:to_i)

K2 = 2 * K
imos = Array.new(K2){Array.new(K, 0)}

N.times do |i|
    x, y, c = gets.chomp.split
    x = x.to_i % K2
    y = y.to_i % K2

    # change condition to
    # 0 <= x < KX, 0 <= y < KY, 'B'
    if c == 'B'
        if y >= K
            x = (x + K) % K2
            y -= K
        end
    else
        if y >= K
            y -= K
        else
            x = (x + K) % K2
        end
    end

    # imos for each rectangle (max-3)
    # ..*##.
    # ##...#
    # ##...#
    
    # top
    imos[x][y] += 1
    imos[x + K][y] -= 1 if x < K

    # left-bottom
    if y > 0
        imos[0][0] += 1
        imos[x][y] += 1
        imos[x][0] -= 1
        imos[0][y] -= 1
    end

    # right-bottom
    if y > 0 && x < K
        imos[x + K][0] += 1
        imos[x + K][y] -= 1
    end
end

(1...K2).each do |i|
    (0...K).each do |j|
        imos[i][j] += imos[i - 1][j]
    end
end

(0...K2).each do |i|
    (1...K).each do |j|
        imos[i][j] += imos[i][j - 1]
    end
end

max = 0
(0...K2).each do |i|
    (0...K).each do |j|
        max = imos[i][j] if max < imos[i][j]
    end
end

print max