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:
#da[i][j]:(0,0)~(i,j)の長方形の和
def da_generate(h,w,a):
    da = [[0]*w for j in range(h)]
    da[0][0] = a[0][0]
    for i in range(1,w):
        da[0][i] = da[0][i-1]+a[0][i]
    for i in range(1,h):
        cnt_w = 0
        for j in range(w):
            cnt_w += a[i][j]
            da[i][j] = da[i-1][j]+cnt_w
    return da

#da_calc(p,q,x,y):(p,q)~(x,y)の長方形の和
def da_calc(da,p,q,x,y):
    if p > x or q > y:
        return 0
    if p == 0 and q == 0:
        return da[x][y]
    if p == 0:
        return da[x][y]-da[x][q-1]
    if q == 0:
        return da[x][y]-da[p-1][y]
    return da[x][y]-da[p-1][y]-da[x][q-1]+da[p-1][q-1]

def examD():
    N, K = LI()
    #y=0,x=0上で、原点とおなじグループかどうか
    colorXY = [[0] * (2*K) for _ in range(2*K)]
    for _ in range(N):
        #原点の色を白(0)とする
        x, y, c = LSI()
        x, y = int(x), int(y)
        if c=="B":
            y +=K
        colorXY[x%(2*K)][y%(2*K)] +=1
    ans = 0
    ansC = N
    da = da_generate(2*K, 2*K, colorXY)
    for i in range(K):
        for j in range(K):
            cur = da_calc(da,0,0,K-i-1,K-j-1) + da_calc(da,K-i,K-j,2*K-i-1,2*K-j-1)\
                + da_calc(da,0,2*K-j,K-i-1,2*K-1) + da_calc(da,2*K-i,0,2*K-1,K-j-1)\
                + da_calc(da,2*K-i,2*K-j,2*K-1,2*K-1)
            ans = max(ans,cur)
            ansC = min(ansC,cur)
    print(max(ans,N-ansC))

import sys,copy,bisect,itertools,heapq,math
from heapq import heappop,heappush,heapify
from collections import Counter,defaultdict,deque
def I(): return int(sys.stdin.readline())
def LI(): return list(map(int,sys.stdin.readline().split()))
def LSI(): return list(map(str,sys.stdin.readline().split()))
def LS(): return sys.stdin.readline().split()
def S(): return sys.stdin.readline().strip()
mod = 10**9 + 7
inf = float('inf')

examD()