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 = sys.stdin.readline

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

E = []
for i in range(n):
    x = int(C[i][0]) % (2*k)
    y = int(C[i][1]) % (2*k)
    if C[i][2] == "W":
        x = (x+k) % (2*k)
    E.append([x, y])
    x = (x+k) % (2*k)
    y = (y+k) % (2*k)
    E.append([x, y])

for i in range(2*n):
    xx = 0
    yy = 0
    x = E[i][0]
    y = E[i][1]

    D[x][y] += 1
    if x+k > 2*k:
        xx = 1
    if y+k > 2*k:
        yy = 1
    if xx == 1 and yy == 1:
        D[0][0] += 1
        D[x-k][0] -= 1
        D[0][y-k] -= 1
        D[x-k][y-k] += 1

        D[x][0] += 1
        D[x][y-k] -= 1

        D[0][y] += 1
        D[x-k][y] -= 1

    elif xx == 1:
        D[0][y] += 1
        D[x-k][y] -= 1
        if y != k:
            D[x][y+k] -= 1
            D[0][y+k] -= 1
            D[x-k][y+k] += 1
    elif yy == 1:
        D[x][0] += 1
        D[x][y-k] -= 1
        if x != k:
            D[x+k][y] -= 1
            D[x+k][0] -= 1
            D[x+k][y-k] += 1
    else:
        if x != k:
            D[x+k][y] -= 1
        if y != k:
            D[x][y+k] -= 1
        if x != k and y != k:
            D[x+k][y+k] += 1
    # print(D)
    
for i in range(1, 2*k):
    D[0][i] += D[0][i-1]
    D[i][0] += D[i-1][0]

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

print(max([max(D[i]) for i in range(2*k)]))