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 PyPy3 (7.3.0) (judge verdict: Time Limit Exceeded)

def calcb(lx,ly,rx,ry):
  if rx<0 or ry<0 or lx>rx or ly>ry:
    return 0
  ret=acum1[ry][rx]
  if lx!=0:
    ret-=acum1[ry][lx-1]
  if ly!=0:
    ret-=acum1[ly-1][rx]
  if lx!=0 and ly!=0:
    ret+=acum1[ly-1][lx-1]
  return ret

def calcw(lx,ly,rx,ry):
  if rx<0 or ry<0 or lx>rx or ly>ry:
    return 0
  ret=acum2[ry][rx]
  if lx!=0:
    ret-=acum2[ry][lx-1]
  if ly!=0:
    ret-=acum2[ly-1][rx]
  if lx!=0 and ly!=0:
    ret+=acum2[ly-1][lx-1]
  return ret

n,k=map(int,input().split())
board1=[[0]*(2*k) for _ in range(2*k)]
board2=[[0]*(2*k) for _ in range(2*k)]
acum1=[[0]*(2*k) for _ in range(2*k)]
acum2=[[0]*(2*k) for _ in range(2*k)]
for _ in range(n):
  x,y,c=input().split()
  x,y=int(x)%(2*k),int(y)%(2*k)
  if c=='B':
    board1[y][x]+=1
  if c=='W':
    board2[y][x]+=1
for i in range(2*k):
  for j in range(2*k):
    acum1[i][j]=board1[i][j]
    acum2[i][j]=board2[i][j]
    if i!=0:
      acum1[i][j]+=acum1[i-1][j]
      acum2[i][j]+=acum2[i-1][j]
    if j!=0:
      acum1[i][j]+=acum1[i][j-1]
      acum2[i][j]+=acum2[i][j-1]
    if i!=0 and j!=0:
      acum1[i][j]-=acum1[i-1][j-1]
      acum2[i][j]-=acum2[i-1][j-1]
ans=0
for i in range(k):
  for j in range(k):
    tmp1=0
    tmp2=0
    lx,ly=i,j
    rx,ry=lx+k-1,ly+k-1
    #1個目のマス
    tmp1+=calcb(lx,ly,rx,ry)
    tmp2+=calcw(lx,ly,rx,ry)
    #左上のマス
    tlx,tly=0,0
    trx,ttry=lx-1,ly-1
    tmp1+=calcb(tlx,tly,trx,ttry)
    tmp2+=calcw(tlx,tly,trx,ttry)
    #右上のマス
    tlx,tly=rx+1,0
    trx,ttry=2*k-1,ly-1
    tmp1+=calcb(tlx,tly,trx,ttry)
    tmp2+=calcw(tlx,tly,trx,ttry)
    #左下のマス
    tlx,tly=0,ry+1
    trx,ttry=lx-1,2*k-1
    tmp1+=calcb(tlx,tly,trx,ttry)
    tmp2+=calcw(tlx,tly,trx,ttry)
    #右下のマス
    tlx,tly=rx+1,ry+1
    trx,ttry=2*k-1,2*k-1
    tmp1+=calcb(tlx,tly,trx,ttry)
    tmp2+=calcw(tlx,tly,trx,ttry)
    #上のマス
    tlx,tly=lx,0
    trx,ttry=rx,ly-1
    tmp1+=calcw(tlx,tly,trx,ttry)
    tmp2+=calcb(tlx,tly,trx,ttry)
    #左のマス
    tlx,tly=0,ly
    trx,ttry=lx-1,ry
    tmp1+=calcw(tlx,tly,trx,ttry)
    tmp2+=calcb(tlx,tly,trx,ttry)
    #右のマス
    tlx,tly=rx+1,ly
    trx,ttry=2*k-1,ry
    tmp1+=calcw(tlx,tly,trx,ttry)
    tmp2+=calcb(tlx,tly,trx,ttry)
    #下のマス
    tlx,tly=lx,ry+1
    trx,ttry=rx,2*k-1
    tmp1+=calcw(tlx,tly,trx,ttry)
    tmp2+=calcb(tlx,tly,trx,ttry)
    ans=max(ans,tmp1,tmp2)
print(ans)