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 LuaJIT (2.0.4) (judge verdict: Accepted)

local n, k = io.read("*n", "*n")
local k2 = k * 2
local rows_per_col = {}
for i = 1, k2 do rows_per_col[i] = {} end
local count_per_row, count_per_row2 = {}, {}
for i = 1, k2 do
  count_per_row[i] = 0
  count_per_row2[i] = 0
end
for i = 1, n do
  local x, y, c = io.read("*n", "*n", "*l")
  if c == " W" then x = x + k end
  x, y = x % k2 + 1, y % k2 + 1
  table.insert(rows_per_col[x], y)
  if x <= k then
    count_per_row[y] = count_per_row[y] + 1
  else
    count_per_row2[y] = count_per_row2[y] + 1
  end
end
local mma = math.max
local function solve()
  local val = 0
  for i = 1, k do
    val = val + count_per_row[i] + count_per_row2[i + k]
  end
  local maxval = val
  for i = 1, k do
    val = val - count_per_row[i] + count_per_row[k + i] - count_per_row2[k + i] + count_per_row2[i]
    maxval = mma(maxval, val)
  end
  for i = 1, k do
    val = val + count_per_row[i] - count_per_row[k + i] + count_per_row2[k + i] - count_per_row2[i]
    maxval = mma(maxval, val)
  end
  return maxval
end
local ret = solve(maxval)
for i = 1, k do
  for j = 1, #rows_per_col[i] do
    count_per_row[rows_per_col[i][j]] = count_per_row[rows_per_col[i][j]] - 1
    count_per_row2[rows_per_col[i][j]] = count_per_row2[rows_per_col[i][j]] + 1
  end
  for j = 1, #rows_per_col[i + k] do
    count_per_row[rows_per_col[i + k][j]] = count_per_row[rows_per_col[i + k][j]] + 1
    count_per_row2[rows_per_col[i + k][j]] = count_per_row2[rows_per_col[i + k][j]] - 1
  end
  ret = mma(ret, solve())
end
for i = 1, k do
  for j = 1, #rows_per_col[i] do
    count_per_row[rows_per_col[i][j]] = count_per_row[rows_per_col[i][j]] + 1
    count_per_row2[rows_per_col[i][j]] = count_per_row2[rows_per_col[i][j]] - 1
  end
  for j = 1, #rows_per_col[i + k] do
    count_per_row[rows_per_col[i + k][j]] = count_per_row[rows_per_col[i + k][j]] - 1
    count_per_row2[rows_per_col[i + k][j]] = count_per_row2[rows_per_col[i + k][j]] + 1
  end
  ret = mma(ret, solve())
end
print(ret)