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<cstdio>
#include<cstring>
#include<iostream>
#include<string>
#include<map>
#include<vector>
#include<queue>
#include<set>
#include<stack>
#include<sstream>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N=100000+10;
struct {
    int x,y;
    char c;
}p[N];
int mabs(int x){
    if(x<0) return x*-1;
    return x;
}
int main() {
    int n,k;
    while(~scanf("%d%d",&n,&k)){
        for(int i=0;i<n;i++){
            cin>>p[i].x>>p[i].y>>p[i].c;
        }
        int _max=0;
        for(int i=0;i<=5;i++)
            for(int j=0;j<=5;j++){
                int sum=0;
                for(int k=0;k<n;k++){
                    int x=mabs(p[k].x+i),y=mabs(p[k].y+j);
                    if((x/3+y/3+2)%2==0){
                        if(p[k].c=='W') sum++;
                    }
                    else {
                        if(p[k].c=='B') sum++;
                    }
                }
                _max=max(_max,sum);
            }

        printf("%d\n",_max);
    }
    return 0;
}