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: Wrong Answer)

fn get_line() -> String {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().to_string()
}

fn reads<T>() -> Vec<T>
    where
        T: std::str::FromStr,
        <T as std::str::FromStr>::Err: std::fmt::Debug {
    get_line().split(' ').map(|x| x.parse().unwrap()).collect()
}

fn main() {
    let nk = reads();
    let mut xs = vec![vec![(0, 0); nk[1] * 2]; nk[1]];

    let k = nk[1];

    for _ in 0..nk[0] {
        let xyc: Vec<usize> = get_line().split(' ').map(|x| {
            if x == "W" {
                1
            }
            else if x == "B" {
                0
            }
            else {
                x.parse().unwrap()
            }
        }).collect();

        let x = xyc[0] % (2 * k);
        let y = xyc[1] % (2 * k);
        let m = xyc[2] == 1;

        let (x, y) = 
        if x >= k && y >= k {
            (x - k, y - k)
        }
        else if x >= k {
            (x - k, y + k)
        }
        else {
            (x, y)
        };

        let (w, b) = xs[x][y];
        xs[x][y] = 
        if m {
            (w + 1, b)
        }
        else {
            (w, b + 1)
        };
    }

    for i in 1..k {
        let (w, b) = xs[i][0];
        let (pw, pb) = xs[i - 1][0];
        xs[i][0] = (w + pw, b + pb);
    }
    for j in 1..(2 * k) {
        let (w, b) = xs[0][j];
        let (pw, pb) = xs[0][j - 1];
        xs[0][j] = (w + pw, b + pb);
    }

    for i in 1..k {
        for j in 1..(2 * k) {
            let (w, b) = xs[i][j];
            let (w1, b1) = xs[i - 1][j];
            let (w2, b2) = xs[i][j - 1];
            let (w3, b3) = xs[i - 1][j - 1];

            xs[i][j] = (w + w1 + w2 - w3, b + b1 + b2 - b3);
        }
    }

    use std::cmp;

    let mut max = -1;

    for i in 0..k {
        for j in 0..k {
            let (w0, b0) = xs[i][j];
            let (w1, b1) = xs[k - 1][j];
            let (w2, b2) = xs[i][k + j];
            let (w3, b3) = xs[k - 1][k + j];
            let (w4, b4) = xs[i][2 * k - 1];
            let (w5, b5) = xs[k - 1][2 * k - 1];

            let v1 = w0 + w1 + w2 * 2 - w3 - w4 + w5 + b0 * 2 - b1 - b2 * 2 + b3 + b4; 
            let v2 = b0 + b1 + b2 * 2 - b3 - b4 + b5 + w0 * 2 - w1 - w2 * 2 + w3 + w4;

            let v = cmp::max(v1 / 2, v2 / 2);
            max = cmp::max(max, v);  
        }
    }

    println!("{}", max);
}