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
from pprint import *
N,K=map(int,input().split())
input=sys.stdin.readline
dpb=[[0 for _ in range(2*K+1)] for _ in range(2*K+1)]
def getmatnum(x1,y1,x2,y2,mat):#左上を(x1,y1)右下を(x2,y2)にしてmatの中の左上~右下の領域の和を算出
    if (x2==0)or(y2==0):
        return 0
    else:
        if (x1==-1):
            x1=0
        if (y1==-1):
            y1=0
        return mat[y2][x2]-mat[y1][x2]-mat[y2][x1]+mat[y1][x1]

for i in range(N):
    x,y,c=input().split()
    x=int(x)
    y=int(y)
    #print(x,y,c)
    if c=="W":
        dpb[(y%(2*K))+1][((x+K)%(2*K))+1]=dpb[(y%(2*K))+1][((x+K)%(2*K))+1]+1
    else:
        dpb[(y % (2*K))+1][(x % (2*K))+1] = dpb[(y % (2*K))+1][(x % (2*K))+1] + 1
#pprint(dpw)
#pprint(dpb)

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

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

ans=0
for i in range(K):
    for j in range(K):
        k=i+K
        l=j+K
        b1=getmatnum(-1,-1,i,j,dpb)+getmatnum(k,-1,2*K,j,dpb)+getmatnum(i,j,k,l,dpb)+getmatnum(-1,l,i,2*K,dpb)+getmatnum(k,l,2*K,2*K,dpb)
        b2=getmatnum(i,-1,k,j,dpb)+getmatnum(-1,j,i,l,dpb)+getmatnum(k,j,2*K,l,dpb)+getmatnum(i,l,k,2*K,dpb)
        ans=max(ans,b1,b2)

        #print(i,j,k,l)
print(ans)
        #
#pprint(dpw)
#pprint(dpb)