Problem:
Score : 500 points

Problem Statement

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:

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.

Sample Input 1

4 3
0 1 W
1 2 W
5 3 B
5 4 B

Sample Output 1

4

He can satisfy all his desires by painting as shown in the example above.

Sample Input 2

2 1000
0 0 B
0 1 W

Sample Output 2

2

Sample Input 3

6 2
1 2 B
2 1 W
2 2 B
1 0 B
0 6 W
4 5 W

Sample Output 3

4

Solution in Rust (1.42.0) (judge verdict: Wrong Answer)

use proconio::input;

fn main() {
    input!(n: usize, k: usize, xyc: [(usize, usize, char); n]);
    let mut geta = 0;
    let mut grid = vec![vec![0; 3 * k + 1]; 2 * k + 1];
    for (x, y, c) in xyc {
        let weight = match c {
            'W' => 1,
            'B' => {
                geta += 1;
                -1
            }
            _ => panic!(),
        };
        let x = x % k;
        let y = y % (2 * k);
        for qx in 0..2 {
            for qy in 0..(if y < k { 2 } else { 1 }) {
                grid[qx * k + x + 1][qy * 2 * k + y + 1] += weight;
            }
        }
    }
    for i in 0..=2 * k {
        for j in 0..=3 * k {
            if 0 < i {
                grid[i][j] += grid[i - 1][j];
            }
            if 0 < j {
                grid[i][j] += grid[i][j - 1];
            }
            if 0 < i && 0 < j {
                grid[i][j] -= grid[i - 1][j - 1];
            }
        }
    }

    println!(
        "{}",
        geta + (0..k)
            .zip(0..2 * k)
            .map(|(i, j)| { grid[i + k][j + k] - grid[i + k][j] - grid[i][j + k] + grid[i][j] })
            .max()
            .unwrap()
    );
}