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)

#![allow(unused_macros)]
#![allow(unused_mut)]
#![allow(non_snake_case)]
#![allow(unused_variables)]
#![allow(unused_imports)]
//
// https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8
//
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)*}
    };
}

macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

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

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>>()
    };

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

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

macro_rules! max {
    ($x: expr) => ($x);
    ($x: expr, $($y: expr),+) => (::std::cmp::max($x, max!($($y),+)));
}

macro_rules! min {
    ($x: expr) => ($x);
    ($x: expr, $($y: expr),+) => (::std::cmp::min($x, min!($($y),+)));
}

// solution
use std::cmp;
use std::collections::{BTreeMap, BTreeSet, HashMap, HashSet, VecDeque};
fn main() {
    input! {
        N: usize,
        K: usize,
        kibo: [(usize, usize, chars); N],
    }
    let mut candidate = vec![ vec![0_isize; 2*K]; 2*K];
    for k in kibo.iter_mut() {
        let mut row = k.0;
        let mut col = k.1;
        let c = k.2[0];
        if c == 'W' {
            row += K;
        }
        row %= 2*K;
        col %= 2*K;

        let i = row.saturating_sub(K-1);
        let j = col.saturating_sub(K-1);
        candidate[i][j] += 1;
        match (i+K < 2*K, j+K < 2*K) {
            (true, true) => {
                candidate[i+K][j+K] -= 1;
                candidate[i+K][j] -= 1;
                candidate[i][j+K] -= 1;
            },
            (true, false) => {
                candidate[i+K][j] -= 1;
            },
            (false, true) => {
                candidate[i][j+K] -= 1;
            },
            (false, false) => (),
        }
    }

    for row in 0..2*K {
        for col in 1..2*K {
            candidate[row][col] += candidate[row][col-1];
        }
    }
    for col in 0..2*K {
        for row in 1..2*K {
            candidate[row][col] += candidate[row-1][col];
        }
    }
    
    let mut max = 0;
    for row in 0..2*K {
        for col in 0..2*K {
            max = cmp::max(max, candidate[row][col]);
        }
    }

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