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 <algorithm>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <iomanip>
#define inf -2147483647
using namespace std;
struct color
{
    int x,y,no;
}s[100000];
int n,k,c;
int check(int x,int y,int no)
{
    for(int i = 1;i < n;i ++)
    {
        int d = s[i].x - x;
        int e = s[i].y - y;
        if(d > 0)d += k - 1;
        else d = - d;
        if(e > 0)e += k - 1;
        else e = -e;
        d = (d / k) % 2;
        e = (e / k) % 2;
        if(s[i].no ^ no && (d ^ e) || s[i].no ^ no == 0 && (d ^ e == 0))return 0;
    }
    return 1;
}
int main()
{
    cin>>n>>k;
    char ch;
    for(int i = 0;i < n;i ++)
    {
        cin>>s[i].x>>s[i].y>>ch;
        s[i].no = ch == 'W' ? 0 : 1;
    }
    for(int i = 0;i < k;i ++)
    {
        for(int j = 0;j < k;j ++)
        {
            if(check(s[0].x + i,s[0].y + j,s[0].no))c ++;
        }
    }
    cout<<c;
}