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
from collections import defaultdict, deque, Counter
from heapq import heappush, heappop, heapify
import math
import bisect
import random
from itertools import permutations, accumulate, combinations, product
import sys
import string
from bisect import bisect_left, bisect_right
from math import factorial, ceil, floor
from operator import mul
from functools import reduce


sys.setrecursionlimit(2147483647)
INF = 10 ** 13
def LI(): return list(map(int, sys.stdin.buffer.readline().split()))
def I(): return int(sys.stdin.buffer.readline())
def LS(): return sys.stdin.buffer.readline().rstrip().decode('utf-8').split()
def S(): return sys.stdin.buffer.readline().rstrip().decode('utf-8')
def IR(n): return [I() for i in range(n)]
def LIR(n): return [LI() for i in range(n)]
def SR(n): return [S() for i in range(n)]
def LSR(n): return [LS() for i in range(n)]
def SRL(n): return [list(S()) for i in range(n)]
def MSRL(n): return [[int(j) for j in list(S())] for i in range(n)]
mod = 1000000007


n, k = LI()
acc = [[0] * 3 * k for _ in range(2 * k)]
for _ in range(n):
    x, y, c = LS()
    x = int(x); y = int(y)
    if c == 'W':
        x += k
    x %= 2 * k
    y %= 2 * k
    acc[y][x] += 1
    for i, j in ((1, 1), (1, -1), (-1, 1), (-1, -1), (0, 2)):
        if 0 <= y + k * i < 2 * k and 0 <= x + k * j < 3 * k:
            acc[y + k * i][x + k * j] += 1



acc = [[0] * (3 * k + 1)] + [list(accumulate([0] + acc[y])) for y in range(2 * k)]
for y in range(2 * k):
    for x in range(3 * k + 1):
        acc[y + 1][x] += acc[y][x]


ans = 0
for y in range(k + 1):
    for x in range(2 * k + 1):
        ans = max(ans, acc[y + k][x + k] - acc[y][x + k] - acc[y + k][x] + acc[y][x])


print(ans)