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:
import sys
def MI(): return map(int,sys.stdin.readline().rstrip().split())
def S(): return sys.stdin.readline().rstrip()

N,K = MI()

black = [[0]*K for _ in range(K)]
white = [[0]*K for _ in range(K)]
for _ in range(N):
    x,y,c = map(str,S().split())
    x,y = int(x),int(y)
    if ((x//K + y//K) % 2 == 0 and c == 'B') or ((x//K + y//K) % 2 == 1 and c == 'W'):
        black[x % K][y % K] += 1
    else:
        white[x % K][y % K] += 1

black_accumulate = [[0]*(K+1)] + [[0]+black[i] for i in range(K)]  # black の2次元累積和
white_accumulate = [[0]*(K+1)] + [[0]+white[i] for i in range(K)]  # white の2次元累積和
for i in range(1,K+1):
    for j in range(1,K+1):
        black_accumulate[i][j] += black_accumulate[i][j-1]
        white_accumulate[i][j] += white_accumulate[i][j-1]
for i in range(1,K+1):
    for j in range(1,K+1):
        black_accumulate[i][j] += black_accumulate[i-1][j]
        white_accumulate[i][j] += white_accumulate[i-1][j]

def f(s,t,u,v):  # black の (s,t]×(u,v] の区間和
    return black_accumulate[t][v]-black_accumulate[t][u]-black_accumulate[s][v]+black_accumulate[s][u]
def g(s,t,u,v):  # white の (s,t]×(u,v] の区間和
    return white_accumulate[t][v]-white_accumulate[t][u]-white_accumulate[s][v]+white_accumulate[s][u]

ans = 0
for i in range(1,K+1):  # 区切る位置
    for j in range(1,K+1):  # 区切る位置
        ans = max(ans,f(0,i,0,j) + f(i,K,j,K) + g(0,i,j,K) + g(i,K,0,j))
        ans = max(ans,g(0,i,0,j) + g(i,K,j,K) + f(0,i,j,K) + f(i,K,0,j))
print(ans)