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_=[list(input().split()) for _ in range(n)]
xyc=[]
for x,y,c in xyc_:
  x=int(x)%(2*k)
  y=int(y)%(2*k)
  if x>=k and y>=k:
    x%=k
    y%=k
  elif x>=k:
    x%=k
    c='W' if c=='B' else 'B'
  elif y>=k:
    y%=k
    c='W' if c=='B' else 'B'
  xyc.append([x,y,c])

wmap=[[0]*k for _ in range(k)]
bmap=[[0]*k for _ in range(k)]
for x,y,c in xyc:
  if c=='B':
    bmap[x][y]+=1
  else:
    wmap[x][y]+=1
for i in range(k):
  for j in range(k-1):
    bmap[i][j+1]+=bmap[i][j]
    wmap[i][j+1]+=wmap[i][j]
for i in range(k):
  for j in range(k-1):
    bmap[j+1][i]+=bmap[j][i]
    wmap[j+1][i]+=wmap[j][i]
#print(xyc)
#for b in bmap:
#  print(b)
#for w in wmap:
#  print(w)

def func(x,y):
  bret,wret=0,0
  bret=bmap[-1][-1]
  wret=wmap[-1][-1]
  if x==0 and y==0:
    pass
  elif x==0:
    bret-=bmap[-1][y-1]
    wret-=wmap[-1][y-1]
  elif y==0:
    bret-=bmap[x-1][-1]
    wret-=wmap[x-1][-1]
  else:
    bret-=bmap[-1][y-1]
    wret-=wmap[-1][y-1]
    bret-=bmap[x-1][-1]
    wret-=wmap[x-1][-1]
    bret+=2*bmap[x-1][y-1]
    wret+=2*wmap[x-1][y-1]
  return max(wmap[-1][-1]-wret+bret,bmap[-1][-1]-bret+wret)
ans=0
for i in range(k):
  for j in range(k):
    ans=max(ans,func(i,j))
print(ans)
