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;

typedef long long ll;

const int INF = 1000000000;


int cal(int x1, int y1, int x2, int y2, vector<vector<int> >& board)
{
  int ret = board[y2][x2] - board[y2][x1] - board[y1][x2] + board[y1][x1];
  return ret;
}

int main()
{
  int N, K;
  scanf("%d %d", &N, &K);
  vector<vector<int> > board(2 * K + 1, vector<int>(2 * K + 1, 0));
  for (int i = 0; i < N; ++i) {
    int x, y;
    char c;
    scanf("%d %d %c", &x, &y, &c);
    if (c == 'W') { x += K; }
    x %= (2 * K);
    y %= (2 * K);
    ++board[y + 1][x + 1];
  }

  for (int y = 1; y <= 2 * K; ++y) {
    for (int x = 1; x <= 2 * K; ++x) {
      board[y][x] += board[y][x - 1] + board[y - 1][x] - board[y - 1][x - 1];
    }
  }
  // fprintf(stderr, "DEBUG: after cumulative summing\n");
  // for (int i = 0; i <= 2 * K; ++i) {
  //   for (int j = 0; j <= 2 * K; ++j) {
  //     fprintf(stderr, "%3d |", board[i][j]);
  //   }
  //   fprintf(stdout, "\n");
  // }
  int ans = 0;
  for (int x = 1; x <= K; ++x) {
    for (int y = 1; y <= K; ++y) {
      int x0 = 0, y0 = 0;
      int x1 = x, y1 = y;
      int x2 = x + K, y2 = y + K;
      int x3 = 2 * K, y3 = 2 * K;
      int cnt = cal(x0, y0, x1, y1, board) +
        cal(x2, y0, x3, y1, board) +
        cal(x1, y1, x2, y2, board) +
        cal(x0, y2, x1, y3, board) +
        cal(x2, y2, x3, y3, board);
      // fprintf(stderr, "DEBUG: (x, y)=(%d, %d), cnt=%d\n", x, y, cnt);
      ans = max(ans, cnt);
      ans = max(ans, N - cnt);
    }
  }
  printf("%d\n", ans);

  return 0;
}