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)

use std::io;

fn main() {
    let mut buf: String = String::new();
    io::stdin().read_line(&mut buf).unwrap();
    let mut iter = buf.trim().split_whitespace();
    let n: i64 = iter.next().unwrap().parse::<i64>().unwrap();
    let k: i64 = iter.next().unwrap().parse::<i64>().unwrap();
    let mut vs: Vec<(i64, i64)> = Vec::new();
    for _ in 0..n {
        let mut buf = String::new();
        io::stdin().read_line(&mut buf).unwrap();
        let mut iter = buf.trim().split_whitespace();
        let x: i64 = iter.next().unwrap().parse::<i64>().unwrap();
        let y: i64 = iter.next().unwrap().parse::<i64>().unwrap();
        let c: char = iter.next().unwrap().chars().next().unwrap();
        if c == 'B' {
            vs.push((x % (2 * k), y % (2 * k)));
        } else {
            vs.push((x % (2 * k), (y + k) % (2 * k)));
        }
    }
    let mut max_c: i32 = 0;
    for i in 0..k {
        for j in 0..k {
            let mut c: i32 = 0;
            for l in 0..(n as usize) {
                let (xx, yy) = vs[l];
                c += if ((i <= xx && xx < k + i ) && (j <= yy && yy < k + j)) ||
                    ((xx < i || k + i <= xx) && (yy < j || k + j <= yy)) { 1 }
                else { 0 };
            }
            max_c = if max_c > c { max_c } else {c};
        }
    }
    println!("{}", max_c);
}