human
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
Using python3 can you solve the prior task?
assistant
#Here is Python code to print the maximum number of desires that can be satisfied at the same time.

#!/usr/bin/env python3
from itertools import chain
import sys
try:
    from typing import List
except ImportError:
    pass

def solve(N: int, K: int, x: "List[int]", y: "List[int]", c: "List[str]"):
    imos = [[0] * (2 * K) for _ in range(2 * K)]

    def sq(x1: int, y1: int, x2: int, y2: int):
        x1 %= 2 * K
        x2 %= 2 * K
        y1 %= 2 * K
        y2 %= 2 * K
        if x1 > x2:
            imos[y1][0] += 1
            imos[y2][0] -= 1
            if y1 > y2:
                imos[0][0] += 1
        if y1 > y2:
            imos[0][x1] += 1
            imos[0][x2] -= 1
        imos[y1][x1] += 1
        imos[y1][x2] -= 1
        imos[y2][x1] -= 1
        imos[y2][x2] += 1

    for xi, yi, ci in zip(x, y, c):
        if ci == "B":
            bx = xi + 1
            by = yi + 1
        else:
            bx = xi - K + 1
            by = yi + 1
        sq(bx, by, bx + K, by + K)
        sq(bx + K, by + K, bx, by)

    # print(imos, file=sys.stderr)

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

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

    # print(imos, file=sys.stderr)
    print(max(chain(*imos)))

def main():
    def iterate_tokens():
        for line in sys.stdin:
            for word in line.split():
                yield word
    tokens = iterate_tokens()
    N = int(next(tokens))  # type: int
    K = int(next(tokens))  # type: int
    x = [int()] * (N)  # type: "List[int]"
    y = [int()] * (N)  # type: "List[int]"
    c = [str()] * (N)  # type: "List[str]"
    for i in range(N):
        x[i] = int(next(tokens))
        y[i] = int(next(tokens))
        c[i] = next(tokens)
    solve(N, K, x, y, c)

if __name__ == '__main__':
    main()