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
class Bit:
    def __init__(self, n):
        self.size = n
        self.tree = [0] * (n + 1)

    def sum(self, i):
        s = 0
        while i > 0:
            s += self.tree[i]
            i -= i & -i
        return s

    def add(self, i, x):
        while i <= self.size:
            self.tree[i] += x
            i += i & -i


N, K = map(int, input().split())
black = []
white = []
for i in range(N):
    x, y, c = map(str, input().split())
    temp = 0
    dx, x = divmod(int(x), K)
    temp += dx
    dy, y = divmod(int(y), K)
    temp += dy
    color = 0 if c == 'B' else 1
    color = (color + temp) % 2
    if color == 0:
        white.append((x, y))
    else:
        black.append((x, y))

white_bit_list = [0]
black_bit_list = [0]
for _ in range(K):
    white_bit = Bit(K)
    black_bit = Bit(K)
    white_bit_list.append(white_bit)
    black_bit_list.append(black_bit)
white_y = [0] * K
black_y = [0] * K
for xy in white:
    x, y = xy
    white_y[y] += 1
    x += 1
    y += 1
    white_bit_list[x].add(y, 1)
for xy in black:
    x, y = xy
    black_y[y] += 1
    x += 1
    y += 1
    black_bit_list[x].add(y, 1)
white_y_cum = [0] * (K + 1)
black_y_cum = [0] * (K + 1)
for i in range(K):
    white_y_cum[i + 1] = white_y_cum[i] + white_y[i]
    black_y_cum[i + 1] = black_y_cum[i] + black_y[i]

ans_best = 0
for dy in range(1, K):
    ul_w = 0
    ur_b = black_y_cum[dy]
    ll_b = 0
    lr_w = white_y_cum[-1] - white_y_cum[dy]
    ans = ul_w + ur_b + ll_b + lr_w
    ans_best = max(ans_best, ans, N - ans)
    for dx in range(1, K):
        ul_w += white_bit_list[dx].sum(dy)
        ur_b -= black_bit_list[dx].sum(dy)
        ll_b += black_bit_list[dx].sum(K) - black_bit_list[dx].sum(dy)
        lr_w -= white_bit_list[dx].sum(K) - white_bit_list[dx].sum(dy)
        ans = ul_w + ur_b + ll_b + lr_w
        ans_best = max(ans_best, ans, N - ans)
l_w = 0
r_b = len(black)
ans = l_w + r_b
ans_best = max(ans_best, ans, N - ans)
for dx in range(1, K):
    l_w += white_bit_list[dx].sum(K)
    r_b -= black_bit_list[dx].sum(K)
    ans = l_w + r_b
    ans_best = max(ans_best, ans, N - ans)

print(ans_best)
