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:

cba927b2484fad94fb5ff7473e9aadef.png

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.

Examples

Input

4 3
0 1 W
1 2 W
5 3 B
5 4 B


Output

4


Input

2 1000
0 0 B
0 1 W


Output

2


Input

6 2
1 2 B
2 1 W
2 2 B
1 0 B
0 6 W
4 5 W


Output

4
# 2K * 2Kの正方形で一周するので、
# (X,Y) -> (X % 2K, Y % 2K)に言い換えることが出来る
# Wの表とBの表を別々で持つ
# それぞれの表で二次元累積和を持ち、範囲に対する得点を求められるようにする
# K * Kの正方形の左下隅を1000 * 1000通り全探索して最大値を求める
# 効率化のため、上記K * Kの正方形が黒である場合と白である場合を同時に計算する
# その領域で見つけた白の個数は、黒を計算するときは「白の総数 - 見つけた白の個数」とすることで白の得点として加算できるえ

import sys
readline = sys.stdin.readline

N,K = map(int,readline().split())
white = [[0] * (2 * K + 1) for i in range(2 * K + 1)]
black = [[0] * (2 * K + 1) for i in range(2 * K + 1)]
# 黒を数えるときは、同時に白を数えて、白全体の数から引く
white_all = 0
black_all = 0
for i in range(N):
  x,y,c = readline().split()
  x,y = int(x) % (2 * K),int(y) % (2 * K)
  if c == "W":
    white[x + 1][y + 1] += 1
    white_all += 1
  else:
    black[x + 1][y + 1] += 1
    black_all += 1
  
for i in range(1, len(white)):
  for j in range(1, len(white[i])):
    white[i][j] += white[i][j - 1]
    
for j in range(1, len(white)):
  for i in range(1, len(white[j])):
    white[i][j] += white[i - 1][j]

for i in range(1, len(black)):
  for j in range(1, len(black[i])):
    black[i][j] += black[i][j - 1]
    
for j in range(1, len(black)):
  for i in range(1, len(black[j])):
    black[i][j] += black[i - 1][j]

cases = ((0,0),(K,K),(-K,-K),(K,-K),(-K,K))
ans = 0
  
for i in range(1, K + 1):
  for j in range(1, K + 1):
    # 基本となる正方形の左下隅・右上は以下の通り
    # (i,j),(i + K - 1, j + K - 1)
    # これに対して、
    # i,jともに +Kしたもの
    # i,jともに -Kしたもの
    # iは+K,jは-Kしたもの
    # iは-K,jは+Kしたもの
    # が考えられる範囲となる
    # なお、最小値は1,1、最大値は2 * K,2 * Kとなる
    # iだけ+Kした正方形は、その正方形の白黒を反転させると計上済みとなる
    black_cnt = 0
    white_cnt = 0
    for c in cases:
      x1 = i + c[0]
      y1 = j + c[1]
      x2 = x1 + K - 1
      y2 = y1 + K - 1
        
      x1 = max(1, x1)
      y1 = max(1, y1)
      x2 = min(2 * K, x2)
      y2 = min(2 * K, y2)
      if x1 > x2 or y1 > y2:
        continue
        
      # その領域の黒と白の数を数える
      black_cnt += black[x2][y2] - black[x2][y1 - 1] - black[x1 - 1][y2] + black[x1 - 1][y1 - 1]
      white_cnt += white[x2][y2] - white[x2][y1 - 1] - white[x1 - 1][y2] + white[x1 - 1][y1 - 1]
      
    # その正方形が黒の場合と白の場合を両方算出し、比較する
    point = max(black_cnt + (white_all - white_cnt),white_cnt + (black_all - black_cnt))
    if ans < point:
      ans = point
    
print(ans)