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)

#[allow(unused_macros)]
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };

    ($next:expr, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

#[allow(unused_macros)]
macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, bytes) => {
        read_value!($next, String).into_bytes()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

#[allow(unused_imports)]
use std::cmp::{max, min};
#[allow(unused_imports)]
use std::collections::{BinaryHeap, HashMap};

#[allow(dead_code)]
type Graph = Vec<Vec<usize>>;

fn main() {
    input! {
        n: usize,
        k: usize,
        p: [(usize, usize, chars); n],
    }

    let k2 = k * 2;

    let p = p
        .into_iter()
        .map(|(x, y, c)| (x % k2, y % k2, c[0]))
        .collect::<Vec<_>>();

    let mut w = vec![vec![0; k2 + 1]; k2 + 1];
    let mut b = vec![vec![0; k2 + 1]; k2 + 1];

    for &(x, y, c) in &p {
        if c as char == 'W' {
            w[y][x] += 1;
        } else {
            b[y][x] += 1;
        }
    }

    let mut wsum = vec![vec![0; k2 + 1]; k2 + 1];
    let mut bsum = vec![vec![0; k2 + 1]; k2 + 1];

    for i in 1..k2 + 1 {
        for j in 1..k2 + 1 {
            wsum[i][j] += wsum[i][j - 1] + w[i - 1][j - 1];
            bsum[i][j] += bsum[i][j - 1] + b[i - 1][j - 1];
        }
    }

    for j in 1..k2 + 1 {
        for i in 1..k2 + 1 {
            wsum[i][j] += wsum[i - 1][j];
            bsum[i][j] += bsum[i - 1][j];
        }
    }

    let mut ans = 0;

    for i in 1..k + 1 {
        for j in 1..k + 1 {
            let mut s1 = 0;
            let mut s2 = 0;
            for ii in 0..3 {
                let i2 = i + ii * k;
                for jj in 0..3 {
                    let j2 = j + jj * k;
                    // println!("{} {} {} {}", i, j, i2, j2);
                    let (add1, add2) = if (ii + jj) % 2 == 0 {
                        (&wsum, &bsum)
                    } else {
                        (&bsum, &wsum)
                    };
                    s1 += add1[min(i2, k2)][min(j2, k2)];
                    s2 += add2[min(i2, k2)][min(j2, k2)];
                    if i2 >= k {
                        s1 -= add1[i2 - k][min(j2, k2)];
                        s2 -= add2[i2 - k][min(j2, k2)];
                    }
                    if j2 >= k {
                        s1 -= add1[min(i2, k2)][j2 - k];
                        s2 -= add2[min(i2, k2)][j2 - k];
                    }
                    if i2 >= k && j2 >= k {
                        s1 += add1[i2 - k][j2 - k];
                        s2 += add2[i2 - k][j2 - k];
                    }
                }
            }
            ans = max(ans, s1);
            ans = max(ans, s2);
        }
    }

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