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 (1.15.1) (judge verdict: Accepted)

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}

fn main() {
    let v = read_vec::<usize>();
    let (n, k) = (v[0], v[1]);
    let mut tiles = vec![vec![0i64; 2 * k]; 2 * k];

    for _ in 0..n {
        let v = read_vec::<String>();
        let mut x: usize = v[0].parse().unwrap();
        let mut y: usize = v[1].parse().unwrap();
        if v[2] == "W" {
            y += k;
        }
        x %= k * 2;
        y %= k * 2;
        if y >= k {
            x = (x + k) % (k * 2);
            y = y - k;
        }

        tiles[x][y] += 1;
        tiles[x][y + k] -= 1;
        if x < k {
            tiles[x + k][y] -= 1;
            tiles[x + k][y + k] += 1;
        } else if x > k {
            tiles[x - k][y] -= 1;
            tiles[x - k][y + k] += 1;
            tiles[0][y] += 1;
            tiles[0][y + k] -= 1;
        }
    }

    for x in 0..2 * k {
        for y in 1..2 * k {
            tiles[x][y] += tiles[x][y - 1];
        }
    }

    for x in 1..2 * k {
        for y in 0..2 * k {
            tiles[x][y] += tiles[x - 1][y];
        }
    }
    let mut ans = 0;
    for x in 0..k * 2 {
        for y in 0..k * 2 {
            ans = std::cmp::max(
                ans,
                tiles[x][y] + tiles[(x + k) % (2 * k)][(y + k) % (2 * k)],
            );
        }
    }
    println!("{}", ans);
}