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: Time Limit Exceeded)

use std::io::prelude::*;

fn input<T>() -> T
where
    T: std::str::FromStr,
{
    let stdin = std::io::stdin();
    let token: String = stdin
        .lock()
        .bytes()
        .map(|c| c.unwrap() as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().unwrap()
}

fn main() {
    let n: usize = input();
    let k: i64 = input();
    let xyc: Vec<(i64, i64, i64)> = (0..n)
        .map(|_| {
            let x = input();
            let y = input();
            let s: String = input();
            let c = if s == "B" { 0 } else { 1 };
            (x, y, c)
        })
        .collect();

    let mut max = 0;
    for c00 in 0..2 {
        for i in 0..k {
            for j in 0..k {
                let mut temp = 0;
                for &(x, y, c) in &xyc {
                    let rx = (x + k - i) % (2 * k);
                    let ry = (y + k - j) % (2 * k);
                    let cxy = if (rx >= k && ry >= k) || (rx < k && ry < k) {
                        c00
                    } else {
                        1 - c00
                    };
                    if cxy == c {
                        temp += 1;
                    }
                }
                max = std::cmp::max(max, temp);
            }
        }
    }
    println!("{}", max);
}