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 2.6.2) (judge verdict: Wrong Answer)

{$R-,S-,Q-,I-,O+}
const KK=1000;
var cond:array[0..1,1..4*KK,1..4*KK] of longint;
    dp:array[0..1,0..4*KK+1,0..4*KK+1] of longint;
    n,k,i,x,y,dx,dy,col,row_sum,max,temp:longint;
    c,d:char;
function sum(c,x1,y1,x2,y2:longint):longint;
begin
  sum:=dp[c,x2,y2]+dp[c,x1-1,y1-1]
        -dp[c,x2,y1-1]-dp[c,x1-1,y2];
end;

begin
  readln(n,k);
  for i:=1 to n do begin
    readln(x,y,c,d);
    x:=x mod (2*k);
    y:=y mod (2*k);
    col:=ord(c='W');
    for dx:=0 to 1 do for dy:=0 to 1 do
      inc(cond[col,x+dx*2*k,y+dy*2*k]);
  end;

  for col:=0 to 1 do
    for x:=1 to 4*k do begin
      row_sum:=0;
      for y:=1 to 4*k do begin
        inc(row_sum,cond[col,x,y]);
        dp[col,x,y]:=dp[col,x-1,y]+row_sum;
      end;
    end;

  for x:=1 to 2*k do for y:=1 to 2*k do begin
    temp:=0;
    inc(temp,sum(0,x+k,y+k,x+2*k-1,y+2*k-1));
    inc(temp,sum(0,x  ,y  ,x+  k-1,y+  k-1));
    inc(temp,sum(1,x+k,y  ,x+2*k-1,y+  k-1));
    inc(temp,sum(1,x  ,y+k,x+  k-1,y+2*k-1));
    if max<temp then max:=temp;
  end;

  writeln(max);
end.