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() {
    input!(n: usize, k: usize, ijc: [(usize, usize, char); n]);
    let mut g = vec![vec![0; k + 1]; k + 1];
    let mut b_count = 0;
    let mut w_count = 0;
    for &(i, j, c) in &ijc {
        g[i % k + 1][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) {
        if 0 < i {
            g[i][j] += g[i - 1][j];
        }
        if 0 < j {
            g[i][j] += g[i][j - 1];
        }
        if 0 < i && 0 < j {
            g[i][j] -= g[i - 1][j - 1];
        }
    }

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