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 Pascal (FPC 3.0.4) (judge verdict: Accepted)

var
	N, K, i, j, x, y, b, nx, ny: integer;
    c: string;
	cnt: array[1..1001, 1..1001] of integer;
begin
	read(N, K);
    for i := 1 to N do begin
    	read(x, y, c);
        if c = ' W' then y := y + k;
        if x div K mod 2 = y div K mod 2 then
        	b := 1
        else
        	b := -1;
        if b = 1 then cnt[1, 1] := cnt[1, 1] + 1;
        nx := x mod K + 2;
        ny := y mod K + 2;
        cnt[nx, 1] := cnt[nx, 1] - b;
        cnt[1, ny] := cnt[1, ny] - b;
        cnt[nx, ny] := cnt[nx, ny] + b * 2;
    end;
    for i := 1 to K + 1 do for j := 1 to K do
    	cnt[i, j + 1] := cnt[i, j + 1] + cnt[i, j];
    for j := 1 to K + 1 do for i := 1 to K do
    	cnt[i + 1, j] := cnt[i + 1, j] + cnt[i, j];
    for i := 1 to K do for j := 1 to K do begin
    	if b < cnt[i, j] then b := cnt[i, j];
    	if b < N - cnt[i, j] then b := N - cnt[i, j];
    end;
    writeln(b);
end.