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: Accepted)

#include <bits/stdc++.h>
using namespace std;

struct data
{
    int u,v;
    char type;
};

data a[100005];
int n,k,res=0,f[1005][2][2];

void Process()
{
    for(int i=0; i<=k; i++)
    {
        for(int j=0; j<k; j++)
        {
            f[j][0][0]=0;
            f[j][0][1]=0;
            f[j][1][0]=0;
            f[j][1][1]=0;
        }
        int Count=0;
        for(int j=1; j<=n; j++)
            if(((a[j].u-i)/k+a[j].v/k)%2==0)
            {
                if(a[j].type=='W')
                {
                    Count++;
                    f[a[j].v%k][0][0]++;
                }
                else
                    f[a[j].v%k][0][1]++;
            }
            else
            {
                if(a[j].type=='B')
                {
                    Count++;
                    f[a[j].v%k][1][1]++;
                }
                else
                    f[a[j].v%k][1][0]++;
            }
        res=max(res,max(Count,n-Count));
        for(int j=0; j<k; j++)
        {
            Count-=f[j][0][0]+f[j][1][1];
            Count+=f[j][0][1]+f[j][1][0];
            res=max(res,max(Count,n-Count));
        }
    }
}

int main()
{
    ios_base::sync_with_stdio(false);
    //freopen("TEST.INP","r",stdin);
    cin>>n>>k;
    for(int i=1; i<=n; i++)
    {
        cin>>a[i].u>>a[i].v>>a[i].type;
        a[i].u+=k;
        a[i].v+=k;
    }
    Process();
    cout<<res;
}