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 (2.4.0) (judge verdict: Wrong Answer)

N, K = map(int, input().split())
K2 = 2*K
R = []

# (x, y) と (x, y+K)の色は必ず異なるので、すべて黒の要求に変換できる。
# 2Kずらすと必ず色が同じになるので、2Kでmodをとってよい。
for i in range(N):
    xi, yi, ci = input().split()
    if ci == "W":
        yi = int(yi) + K

    xi, yi = int(xi), int(yi)
    R.append([yi % K2, xi % K2])


class Ruisekiwa2D:
    def __init__(self, h, w):
        self.H = h
        self.W = w
        self.G = [[0] * w for _ in range(h)]

    def load(self, points):
        # 問題によってイメージしやすいよう書き換えてください
        for x, y in points:
            self.G[x][y] += 1

        self.calc()

    def load_imos(self, points):
        # pointsは[[左上h, 左上w, 右下h, 右下w], ...]であること
        for i1, j1, i2, j2 in points:
            self.G[i1][j1] += 1  # 左上
            if j2 + 1 < self.W:  # 右上
                self.G[i1][j2+1] -= 1
            if i2 + 1 < self.H:
                self.G[i2+1][j1] -= 1  # 左下
            if j2 + 1 < self.W and i2 + 1 < self.H:  # 右下
                self.G[i2+1][j2+1] += 1

        self.calc()

    def calc(self):
        for y in range(self.W):
            for x in range(1, self.H):
                self.G[x][y] += self.G[x-1][y]

        for y in range(1, self.W):
            for x in range(self.H):
                self.G[x][y] += self.G[x][y-1]

    def get(self, i1, j1, i2, j2):
        if i1 > i2 or j1 > j2 or any(map(lambda x: x < 0, [i1, j1, i2, j2])):
            return 0
        elif i1 > 0 and j1 > 0:
            return self.G[i2][j2] - self.G[i1-1][j2] - self.G[i2][j1-1] + self.G[i1-1][j1-1]
        elif i1 <= 0 < j1:
            return self.G[i2][j2] - self.G[i2][j1-1]
        elif j1 <= 0 < i1:
            return self.G[i2][j2] - self.G[i1 - 1][j2]
        else:
            return self.G[i2][j2]


R2D = Ruisekiwa2D(K2, K2)
R2D.load(R)

ans = 0
for i in range(K):
    for j in range(K):
        tmp = 0
        tmp += R2D.get(0, 0, i-1, j-1)  # 左上
        tmp += R2D.get(0, j+K, i-1, K2-1)  # 右上
        tmp += R2D.get(i+K, j+K, K2-1, K2-1)  # 右下
        tmp += R2D.get(i+K, 0, K2-1, j-1)  # 左下
        tmp += R2D.get(i, j, i+K-1, j+K-1)  # 真ん中

        ans = max(ans, tmp)
print(ans)