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
# time complexity: O(N + K^2)
# space complexity: O(K^2)

import sys
fin = sys.stdin.readline
from collections import defaultdict
from pprint import pprint

def sum_range_2D(cuml_mat, coord_left_up, Kx, Ky):
    if Kx <= 0 or Ky <= 0:
        return 0
    x, y = coord_left_up
    if x - 1 >= 0 and y - 1 >= 0:
        left_up = cuml_mat[x - 1][y - 1]
    else:
        left_up = 0
    if x - 1 >= 0 and y + Ky - 1 >= 0:
        right_up = cuml_mat[x - 1][y + Ky - 1]
    else:
        right_up = 0
    if x + Kx - 1 >= 0 and y - 1 >= 0:
        left_down = cuml_mat[x + Kx - 1][y - 1]
    else:
        left_down = 0
    right_down = cuml_mat[min(len(cuml_mat) - 1, x + Kx - 1)][min(len(cuml_mat) - 1, y + Ky - 1)]
    return right_down - right_up - left_down + left_up


N, K = [int(elem) for elem in fin().split()]
black_desires = []
for _ in range(N):
    x, y, c = fin().split()
    x = int(x) % (2 * K)
    y = int(y) % (2 * K) if c == 'B' else (int(y) + K) % (2 * K)
    black_desires.append((x, y))

cuml_2D = [[0] * (2*K) for _ in range(2*K)]
for x, y in black_desires:
    cuml_2D[x][y] += 1

for i in range(2 * K):
    for j in range(1, 2 * K):
        cuml_2D[i][j] += cuml_2D[i][j - 1]
for j in range(2 * K):
    for i in range(1, 2 * K):
        cuml_2D[i][j] += cuml_2D[i - 1][j]

max_num = 0
for i in range(K + 1):
    for j in range(K):
        current_num = sum_range_2D(cuml_2D, (i, j), K, K) \
                    + sum_range_2D(cuml_2D, (0, 0), i, j) \
                    + sum_range_2D(cuml_2D, (0, j + K), i, K - j) \
                    + sum_range_2D(cuml_2D, (i + K, 0), K - i, j) \
                    + sum_range_2D(cuml_2D, (i + K, j + K), K - i, K - j)
        max_num = max(max_num, current_num)


for i in range(K + 1):
    for j in range(K, 2 * K):
        current_num = sum_range_2D(cuml_2D, (i, j), K, 2 * K - j) \
                    + sum_range_2D(cuml_2D, (0, j - K), i, K) \
                    + sum_range_2D(cuml_2D, (i + K, j - K), K - i, K) \
                    + sum_range_2D(cuml_2D, (i, 0), K, j - K)
        max_num = max(max_num, current_num)

print(max_num)