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 Go (1.6) (judge verdict: Wrong Answer)

// Package main provides
//
// File:  d.go
// Author: ymiyamoto
//
// Created on Wed Jan  2 23:01:59 2019
//
package main

import (
	"fmt"
	"sort"
)

func max(slice ...int) int {
	sort.Ints(slice)
	return slice[len(slice)-1]
}

func main() {
	var N, K int
	fmt.Scan(&N, &K)

	k := 2 * K
	board := make([][]int, k)
	for i := range board {
		board[i] = make([]int, k)
	}

	for i := 0; i < N; i++ {
		var x, y int
		var c string
		fmt.Scan(&x, &y, &c)
		if c == "B" {
			x += K
		}
		board[y%k][x%k]++
	}

	for i := range board {
		for j := range board[i] {
			if j+1 < len(board[i]) {
				board[i][j+1] += board[i][j]
			}
		}
	}

	for i := range board {
		for j := range board[i] {
			if i+1 < len(board[i]) {
				board[i+1][j] += board[i][j]
			}
		}
	}

	ans := 0
	for i := 0; i < len(board)-K; i++ {
		for j := 0; j < len(board[i])-K; j++ {
			count := 0
			if i > 0 && j > 0 {
				count += board[i+K-1][j+K-1] - board[i+K-1][j-1] - board[i-1][j+K-1] + board[i-1][j-1] //middle
			} else {
				count += board[i+K-1][j+K-1]
			}

			if i > 0 && j > 0 {
				count += board[i-1][j-1] //lower left
			}

			if j > 0 {
				count += board[k-1][j-1] - board[i+K-1][j-1] //upper left
			}

			if i > 0 {
				count += board[i-1][k-1] - board[i-1][j+K-1] //lower right
			}

			if i+K-1 < len(board) && j+K-1 < len(board[i]) {
				count += board[k-1][k-1] - board[k-1][j+K-1] - board[i+K-1][k-1] + board[i+K-1][j+K-1] //upper right
			}

			ans = max(ans, count)
		}
	}

	fmt.Println(ans)
}