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 <iostream>
#include<iomanip>
#include <map>
#include <algorithm>
#include <vector>
#include <string>
#include <utility>
#include <queue>
#define INF 1e9+7
#define rep(i,n) for(int i=0;i<n;i++)
#define NO cout<<"NO"<<endl;
#define YES cout<< "YES"<<endl;
#define No cout<< "No"<<endl;
#define Yes cout<< "Yes"<<endl;
#define all(a) a.begin(),a.end()
#define P pair<int,int>
#define space ' '
using namespace std;
typedef long long int ll;
typedef unsigned long long int ull;

//(int)'a'は97 (int)'A'は65 (int)'1'は49
//おまじない

const int maxn = 100000;
const int maxk = 2000;

int n,k;
int ans=0;
int x[maxn],y[maxn];
char c[maxn+1];
int d[2*maxk][2*maxk]={0};

int main(){
    
    cin >> n>>k;
    rep(i,n){
        cin >> x[i]>>y[i]>>c[i];
        if(c[i]=='B'){
            x[i]+=k;
            c[i]='W';
        }
        x[i]=x[i]%(2*k);
        y[i]=y[i]%(2*k);
        d[y[i]][x[i]]++;
    }
    
    rep(i,2*k){
        rep(j,2*k){
            if(i==0&&j==0){
                continue;
            }else if(i==0){
                d[i][j]+=d[i][j-1];
            }else if(j==0){
                d[i][j]+=d[i-1][j];
            }else{
                d[i][j]+=d[i][j-1]+d[i-1][j]-d[i-1][j-1];
            }
        }
    }
    
    rep(y,k){
        rep(x,k){
            int can = 0;
            if(y==0&&x==0){
                can += d[y+k-1][x+k-1];
            }else if(y==0){
                can += d[y+k-1][x+k-1]-d[y+k-1][x-1];
            }else if(x==0){
                can += d[y+k-1][x+k-1]-d[y-1][x+k-1];
            }else{
                can += d[y+k-1][x+k-1]-d[y+k-1][x-1]-d[y-1][x+k-1]+d[y-1][x-1];
            }
            
            if(x!=0&&y!=0){
                can += d[y-1][x-1];
            }
            if(x!=0){
                can += d[2*k-1][x-1]-d[y+k-1][x-1];
            }
            if(y!=0){
                can += d[y-1][2*k-1]-d[y-1][x+k-1];
            }
            can += d[2*k-1][2*k-1]-d[2*k-1][x+k-1]-d[y+k-1][2*k-1]+d[y+k-1][x+k-1];
            ans = max(ans,max(can,n-can));
            
        }
    }
    
    
    cout << ans<<endl;
    
    return 0;
}