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)

#[allow(unused_imports)]
use {
    proconio::{fastout, input, marker::*},
    std::cmp::*,
    std::collections::*,
    std::ops::*,
};

fn main() {
    input! {
        n: usize,
        k: usize,
        xyc: [(usize, usize, char); n]
    }
    let mut map = vec![vec![0; 2 * k + 1]; k + 1];
    for (x, y, c) in xyc {
        let mut x = x;
        let mut y = y;
        if c == 'W' {
            x += k;
        }
        x %= 2 * k;
        y %= 2 * k;
        if x >= k {
            x -= k;
            y += k;
            y %= 2 * k;
        }
        map[x + 1][y + 1] += 1;
    }
    for x in 0..k {
        for y in 0..2 * k {
            map[x + 1][y + 1] += map[x + 1][y] + map[x][y + 1] - map[x][y];
        }
    }

    let mut ans = 0;
    for x in 0..=k {
        let sum = map[k][k] + map[x][2 * k] - 2 * map[x][k];
        ans = max(ans, max(n - sum, sum));
    }
    for y in 1..=k {
        let sum = map[k][y + k] - map[k][y];
        ans = max(ans, max(sum, n - sum));
        for x in 1..=k {
            let sum = map[k][y + k] - map[k][y] + map[x][2 * k] + 2 * map[x][y] - 2 * map[x][y + k];
            ans = max(ans, max(sum, n - sum));
        }
    }
    println!("{}", ans);
}