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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <string>
#include <cmath>
using namespace std;

int main() {
    long N, K;
    cin >> N >> K;
    long in_x[N] , in_y[N];
    long in_xa , in_ya;
    int in_c[N];
    for(int i = 0; i < N ; i++)
    {
        string color;
        cin >> in_xa >> in_ya >> color;
        in_x[i] = in_xa % (2*K);
        in_y[i] = in_ya % (2*K);

        if(color == "W")
        {
            in_c[i] = 0;
        }
        else
        {
            in_c[i] = 1;
        }
    }

    int count = 0;
    int max_count = 0;
    for(int x = 0; x < K ; x++)
    {
        for(int y = 0; y < K ; y++)
        {
            for (int i = 0; i < N; i++)
            {
                count += ((in_x[i] + x) / K + (in_y[i] + y) / K) % 2 == in_c[i];
//                cout << (in_y[i] - y)/K << endl;
            }
            if(count > max_count)
            {
                max_count = count;

            }
            else if(N - count > max_count)
            {
                max_count = N - count;
            }
            if(max_count == N)
            {
                cout << max_count << endl;
                return 0;
            }
            count = 0;
        }
    }
    cout << max_count << endl;

    return 0;
}