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
import sys
input = lambda : sys.stdin.readline().rstrip()
sys.setrecursionlimit(max(1000, 10**9))
write = lambda x: sys.stdout.write(x+"\n")



n,k = list(map(int, input().split()))
v = [[0]*(2*k) for _ in range(2*k)]

for i in range(n):
    x,y,c = input().split()
    x = int(x)
    y = int(y)
    if c=="B":
        x += k
    x %= 2*k
    y %= 2*k
    v[x][y] += 1
#     v[x][y+2*k] += 1
#     v[x+2*k][y] += 1
#     v[x+2*k][y+2*k] += 1
def cumsum2(v):
    # v: h*wの2次元リスト
    h,w = len(v), len(v[0])
    v.insert(0, [0]*w)
    for l in v:
        l.insert(0, 0)
    for i in range(1,h+1):
        for j in range(1,w+1):
            v[i][j] = v[i-1][j] + v[i][j-1] - v[i-1][j-1] + v[i][j]
    return v
cumsum2(v)

def sub(c, i, j):
    i %= 2*k
    j %= 2*k
    l = []
    if i<=k and j<=k:
        l = [[i,i+k,j,j+k]]
    elif i>k and j<=k:
        l = [[i,2*k,j,j+k], [0,i-k,j,j+k]]
    elif i<=k and j>k:
        l = [[i,i+k,j,2*k], [i,i+k,0,j-k]]
    else:
        l = [[i,2*k,j,2*k], [0,i-k,j,2*k], [i,2*k,0,j-k], [0,i-k,0,j-k]]
    return sum((c[z][w] - c[z][y] - c[x][w] + c[x][y]) for x,z,y,w in l)
ans = -1
for i in range(2*k):
    for j in range(2*k-i):
        val0 = sub(v, i,j) + sub(v, i+k, j+k)
#         val0 = v[i+k][j+k]-v[i+k][j]-v[i][j+k]+v[i][j] + v[i+2*k][j+2*k]-v[i+2*k][j+k]-v[i+k][j+2*k]+v[i+k][j+k]
#         val1 = v[i+2*k][j+k]-v[i+2*k][j]-v[i+k][j+k]+v[i+k][j] + v[i+k][j+2*k]-v[i+k][j+k]-v[i][j+2*k]+v[i][j+k]
        ans = max(ans, val0)
print(ans)