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
from itertools import accumulate
import sys
input = sys.stdin.readline

N, K = map(int, input().split())
black = [[0] * (2*K+1) for _ in range(2*K+1)]
white = [[0] * (2*K+1) for _ in range(2*K+1)]
for _ in range(N):
    x, y, c = input().split()
    x = int(x) % (2*K) + 1
    y = int(y) % (2*K) + 1
    if c == 'B':
        black[x][y] += 1
    else:
        white[x][y] += 1

for i in range(1, 2*K+1):
    black[i] = list(accumulate(black[i]))
    white[i] = list(accumulate(white[i]))
    for j in range(2*K+1):
        black[i][j] += black[i-1][j]
        white[i][j] += white[i-1][j]

ans = 0
for i in range(K):
    for j in range(K):
        cnt = 0
        for k in [-1, 0, 1]:
            for l in [-1, 0, 1]:
                r = i + K*k
                c = j + K*l
                color = black if (k + l) % 2 == 0 else white
                s1 = color[min(r+K, 2*K)][min(c+K, 2*K)]
                s2 = color[min(r+K, 2*K)][max(c, 0)]
                s3 = color[max(r, 0)][min(c+K, 2*K)]
                s4 = color[max(r, 0)][max(c, 0)]
                cnt += s1 - s2 - s3 + s4
        ans = max(ans, cnt, N - cnt)
print(ans)
