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;
#define rep(i,N) for(int i=0,i##_max=(N);i<i##_max;++i)
#define repp(i,l,r) for(int i=(l),i##_max=(r);i<i##_max;++i)
#define per(i,N) for(int i=(N)-1;i>=0;--i)
#define perr(i,l,r) for(int i=r-1,i##_min(l);i>=i##_min;--i)
#define all(arr) (arr).begin(), (arr).end()
#define SP << " " <<
#define SPF << " "
#define SPEEDUP cin.tie(0);ios::sync_with_stdio(false);
#define MAX_I INT_MAX //1e9
#define MIN_I INT_MIN //-1e9
#define MAX_UI UINT_MAX //1e9
#define MAX_LL LLONG_MAX //1e18
#define MIN_LL LLONG_MIN //-1e18
#define MAX_ULL ULLONG_MAX //1e19
  typedef long long ll;
  typedef pair<int,int> PII;
  typedef pair<char,char> PCC;
  typedef pair<ll,ll> PLL;
  typedef pair<char,int> PCI;
  typedef pair<int,char> PIC;
  typedef pair<ll,int> PLI;
  typedef pair<int,ll> PIL; 
  typedef pair<ll,char> PLC; 
  typedef pair<char,ll> PCL; 

inline void YesNo(bool b){ cout << (b?"Yes" : "No") << endl;}
inline void YESNO(bool b){ cout << (b?"YES" : "NO") << endl;}
inline void Yay(bool b){ cout << (b?"Yay!" : ":(") << endl;}


int N,K;
int table[2005][2005];

int main(void){
  SPEEDUP
  cout << setprecision(15);
  cin >> N >> K;
  const int L = 2*K;
  rep(i,N){
    int x,y;cin >> x >> y;
    char c; cin >> c;
    x%=L;
    y%=L;
    int black = 1;
    if(c=='W')black^=1;
    if(x>=K)black^=1;
    if(y>=K)black^=1;
    x%=K;
    y%=K;
    if(black){
      ++table[0][0];
      --table[0][y+1];
      --table[x+1][0];
      table[x+1][y+1]+=2;
    }else{
      ++table[0][y+1];
      ++table[x+1][0];
      table[x+1][y+1]-=2;
    }
  }
  rep(x,K)repp(y,1,K)table[x][y] += table[x][y-1];
  rep(y,K)repp(x,1,K)table[x][y] += table[x-1][y];
  int ans = 0;
  rep(x,K)rep(y,K){
    int t = N-table[x][y] > table[x][y] ? N-table[x][y]:table[x][y];
    ans = t > ans ? t : ans;
    //ans= max(ans,max(table[x][y],N-table[x][y]));
  }
  cout << ans << endl;
  return 0;
}