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
N,K = map(int,input().split())
XYC = [[int(x)%(2*K),int(y)%(2*K),int(c=="B")] for x,y,c in [input().split() for _ in [0]*N]]

score = [[[0]*(3*K+1) for _ in [0]*(3*K+1)] for _ in [0]*2]
for x,y,c in XYC:
    x+=1
    y+=1
    score[c][y][x] += 1
    if x<K: score[c][y][x+2*K] += 1
    if y<K: score[c][y+2*K][x] += 1
    if x<K and y<K: score[c][y+2*K][x+2*K] += 1

S = score
for c in range(2):
    for j in range(3*K):
        for i in range(3*K):
            S[c][i+1][j+1] += S[c][i][j+1]
    for i in range(3*K):
        for j in range(3*K):
            S[c][i+1][j+1] += S[c][i+1][j]

def get_sum(c,si,sj,i,j):
    i+=1
    j+=1
    return S[c][i][j] + S[c][si][sj] - S[c][i][sj] - S[c][si][j]

ans = 0
for i in range(K):
    for j in range(2*K):
        tmp = 0
        tmp += get_sum(0,i,j,i+K-1,j+K-1)
        tmp += get_sum(1,i+K,j,i+2*K-1,j+K-1)
        j = (j+K)%(2*K)
        tmp += get_sum(1,i,j,i+K-1,j+K-1)
        tmp += get_sum(0,i+K,j,i+2*K-1,j+K-1)
        ans = max(ans,tmp)
print(ans)