Problem: 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

Solution (in PYTHON3):

Solution:
class Accumulate2D: #2次元累積和
    def __init__(self, a):#a=graph
        self.height = len(a)
        self.width = len(a[0])
        self.acc = self.accumulate(a)

    def accumulate(self,a):
        b = [[0]*(self.width+1) for i in range(self.height+1)]
        b[1][1] = a[0][0]
        for i in range(self.height):
            for j in range(self.width):
                b[i+1][j+1] = b[i+1][j]+b[i][j+1]-b[i][j]+a[i][j]
        return b

    def sum(self, p, q):
        y1,x1 = p
        y2,x2 = q
        b = self.acc
        if y1>y2 or x1>x2:
            return 0
        else:
            return b[y2+1][x2+1]-b[y2+1][x1]-b[y1][x2+1]+b[y1][x1]


n,k = map(int, input().split())

kk = k*2

z = [[0]*kk for i in range(kk)]#2k*2kのgraph

for i in range(n):
    x,y,c = input().split()
    x = (int(x) + (k if c=='W'else 0)) % kk#Wの場合はy+kにして、k*kマスへ変換
    y = int(y) % kk#k*kマスへ変換
    z[x][y]+=1

z = Accumulate2D(z)

ans = 0
for i in range(k):
    for j in range(k):#i,jについて全探索
        a = z.sum([0,0],[i-1,j-1])\
            + z.sum([i,j],[i+k-1,j+k-1])\
            + z.sum([0,j+k],[i-1,kk-1])\
            + z.sum([i+k,0],[kk-1,j-1])\
            + z.sum([i+k,j+k],[kk-1,kk-1])
        ans = max(ans,a,n-a)

print(ans)