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 <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef long long LL;
typedef pair<int,int> ii;
typedef vector<ii> vii;
typedef vector<int> vi;
typedef pair<ll,int> li;
#ifdef DEBUG
    #define cek(x) cout<<x
#else
    #define cek(x) if(false){}
#endif // DEBUG

#define fi first
#define se second
#define INF 1000000000
#define INFLL 1000000000000000000LL
#define EPS 1e-9
#define PI acos(-1.0)
#define pb push_back
#define TC() while(tc--)
#define FOR(i,n) for(int i=0;i<n;i++)
#define FORN(i,n) for(int i=0;i<=n;i++)
#define REP(i,a,b,c) for(int i=a;i<b;i+=c)
#define REPN(i,a,b,c) for(int i=a;i<=b;i+=c)
#define reset(a,b) memset(a,b,sizeof(a))

int cum[4010][4010], c[4010][4010];
int n,k,k2;

void calc() {
  FOR(i,2 * k2) {
    FOR(j,2 * k2) {
      cum[i][j] = c[i][j];
      if(i) {
        cum[i][j] += cum[i - 1][j];
      }

      if(j) {
        cum[i][j] += cum[i][j - 1];
      }

      if(i && j) {
        cum[i][j] -= cum[i - 1][j - 1];
      }
    }
  }
}

int ct(int x, int y, int x1, int y1) {
  int tot = cum[x1][y1];
  if(x > 0) {
    tot -= cum[x - 1][y1];
  }

  if(y > 0) {
    tot -= cum[x1][y - 1];
  }

  if (x > 0 && y > 0) {
    tot += cum[x - 1][y - 1];
  }

  return tot;
}

int main(void) {
  std::ios::sync_with_stdio(false);
  reset(cum, 0);
  reset(c, 0);
  cin >> n >> k;
  k2 = 2 * k;

  FOR(i,n) {
    int x,y;
    string col;
    cin >> x >> y >> col;
    x %= k2;
    y %= k2;
    if(col[0] == 'B') {
      c[x][y]++;
      c[x + k2][y + k2]++;
    } else {
      x += k;
      x %= k2;
      c[x][y]++;
      c[x + k2][y + k2]++;
    }
  }

  calc();

  int maxi = 0;

  FOR(i,k2) {
    FOR(j,k2) {
      int tot = ct(i, j, i + k - 1, j + k - 1);
      tot += ct(i + k, j + k, i + k2 - 1, j + k2 - 1);
      // cout << tot << " ";
//       if(i - 1 >= 0 && j - 1 >= 0) {
//         tot += ct(0, 0, i - 1, j - 1);
//       }
//       // cout << tot << " ";
//       if(i - 1 >= 0 && j + k < k2) {
//         tot += ct(0, j + k, i - 1, k2 - 1);
//       }
// // cout << tot << " ";
//       if(i + k < k2 && j - 1 >= 0) {
//         tot += ct(i + k, 0, k2 - 1, j - 1);
//       }
// // cout << tot << " ";
//       if(i + k < k2 && j + k < k2) {
//         tot += ct(i + k, j + k, k2 - 1, k2 - 1);
//       }
// cout << tot << endl;
      maxi = max(maxi, tot);
    }
  }

  assert(maxi <= n);
  cout << maxi << endl;
  return 0;
}