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
import sys
import heapq, math
from itertools import zip_longest, permutations, combinations, combinations_with_replacement
from itertools import accumulate, dropwhile, takewhile, groupby
from functools import lru_cache
from copy import deepcopy

N, K = map(int, input().split())
X, Y, C = [[0] * N for _ in range(3)]

mem = [[0] * (4 * K + 1) for _ in range(4 * K + 1)]

for i in range(N):
    S = input().split()
    X[i], Y[i], C[i] = int(S[0]), int(S[1]), S[2]
    x, y = X[i] % (2 * K) + 1, (Y[i] + (K if C[i] == "W" else 0)) % (2 * K) + 1
    mem[x][y] += 1
    mem[x + K][y] -= 1
    mem[x][y + K] -= 1
    mem[x + K][y + K] += 1

for h in range(1, 4 * K + 1):
    for w in range(1, 4 * K + 1):
        mem[h][w] += mem[h - 1][w] + mem[h][w - 1] - mem[h - 1][w - 1]

ans = 0
for h in range(K + 1):
    for w in range(K + 1):
        s = mem[h][w] + mem[h + 2 * K][w] + mem[h][w + 2 * K] + mem[h + K][w + K] + mem[h + 2 * K][w + 2 * K]
        ans = max(ans, s)

for h in range(K + 1):
    for w in range(K, 2 * K + 1):
        s = mem[h][w] + mem[h + 2 * K][w] + mem[h + K][w - K] + mem[h + K][w + K]
        ans = max(ans, s)


print(ans)