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

"""
@auther: SPD_9X2
https://atcoder.jp/contests/abc086/tasks/arc089_b

考察：

まず、x,y座標を2Kで割っておいてよい。

黒白
白黒　で考え、左下(0,0)で開始。

縦軸を固定し、左に動かしていくことを考える。
すると、現在　x%K == 0　となる部分にいる希望のみ、達成・非達成が反転する

初期の達成数を記録しておき、各列で達成・非達成がいくつあるのかをリストに記録
ずらすと、現在の達成数 += 新たに達成する要求 - 非達成になる要求
となる。2*K回操作して元に戻す

縦軸の移動は各行について、どの列に要求があるのかをおいておき、その部分の達成・非達成数を反転する
各要求で高々2回しかチェックしないため、O(N + K^2)で処理できる
"""




N,K = map(int,input().split())

cansub = [ [] for i in range(2*K) ]
notsub = [ [] for i in range(2*K) ]

notlis = [0] * (2*K)
canlis = [0] * (2*K)

for i in range(N):

    x,y,c = input().split()
    x = int(x)
    y = int(y)
    x %= (2*K)
    y %= (2*K)

    if (x >= K and y >= K) or (x < K and y < K):
        nc = "W"
    else:
        nc = "B"

    if nc == c:
        cansub[x].append(y)
        canlis[y] += 1
    else:
        notsub[x].append(y)
        notlis[y] += 1



ans = 0

for i in range(2*K):

    ncan = sum(canlis)

    for j in range(2*K):

        j %= (2*K)

        ans = max(ans,ncan)
        ncan += notlis[j]
        ncan -= canlis[j]

        t = notlis[j]
        notlis[j] = canlis[j]
        canlis[j] = t

        j += K
        j %= (2*K)

        ncan += notlis[j]
        ncan -= canlis[j]

        t = notlis[j]
        notlis[j] = canlis[j]
        canlis[j] = t


    for k in cansub[i]:
        notlis[k] += 1
        canlis[k] -= 1
    for k in notsub[i]:
        notlis[k] -= 1
        canlis[k] += 1

    t = cansub[i]
    cansub[i] = notsub[i]
    notsub[i] = t

    i += K
    i %= (2*K)

    for k in cansub[i]:
        notlis[k] += 1
        canlis[k] -= 1
    for k in notsub[i]:
        notlis[k] -= 1
        canlis[k] += 1

    t = cansub[i]
    cansub[i] = notsub[i]
    notsub[i] = t

print (ans)