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 (1.42.0) (judge verdict: Accepted)

use itertools::{iproduct, Itertools};
use proconio::input;

fn main() {
    unsafe {
        input!(n: usize, k: usize, ijc: [(usize, usize, char); n]);
        let mut g = [[0; 1001]; 1001];
        let mut b_count = 0;
        let mut w_count = 0;
        for &(i, j, c) in &ijc {
            *g.get_unchecked_mut(i % k + 1).get_unchecked_mut(j % k + 1) +=
                if ((i / k + j / k) % 2 == 1) ^ (c == 'W') {
                    w_count += 1;
                    1
                } else {
                    b_count += 1;
                    -1
                };
        }
        for (i, j) in iproduct!(0..k, 0..k) {
            *g.get_unchecked_mut(i + 1).get_unchecked_mut(j + 1) +=
                g.get_unchecked(i + 1).get_unchecked(j) + g.get_unchecked(i).get_unchecked(j + 1)
                    - g.get_unchecked(i).get_unchecked(j);
        }

        let (min, max): (i32, i32) = iproduct!(0..k, 0..k)
            .map(|(i, j)| {
                g.get_unchecked(k).get_unchecked(k)
                    - g.get_unchecked(i).get_unchecked(k)
                    - g.get_unchecked(k).get_unchecked(j)
                    + 2 * g.get_unchecked(i).get_unchecked(j)
            })
            .minmax()
            .into_option()
            .unwrap();
        println!("{}", std::cmp::max(-min + w_count, max + b_count));
    }
}