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 <vector>
using namespace std;
int score[2001][2001];
void imos(int x1, int y1, int x2, int y2) {
  score[x1][y1]++;
  score[x1][y2]--;
  score[x2][y1]--;
  score[x2][y2]++;
}
int main(void) {
  int n, k;
  cin >> n >> k;
  vector<int> x(n), y(n);
  char color;
  for (int i = 0; i < n; i++) {
    cin >> x[i] >> y[i] >> color;
    if (color == 'B') x[i] += k;
    x[i] %= 2 * k;
    y[i] %= 2 * k;
  }
  for (int i = 0; i < n; i++) {
    if (x[i] < k == y[i] < k) {
      x[i] = x[i] % k;
      y[i] = y[i] % k;
      imos(0, 0, x[i] + 1, y[i] + 1);
      imos(x[i] + 1, y[i] + 1, x[i] + k + 1, y[i] + k + 1);
      imos(x[i] + k + 1, 0, 2 * k, y[i] + 1);
      imos(0, y[i] + k + 1, x[i] + 1, 2 * k);
      imos(x[i] + k + 1, y[i] + k + 1, 2 * k, 2 * k);
    } else {
      x[i] = x[i] % k;
      y[i] = y[i] % k;
      imos(x[i] + 1, 0, x[i] + k + 1, y[i] + 1);
      imos(0, y[i] + 1, x[i] + 1, y[i] + k + 1);
      imos(x[i] + k + 1, y[i] + 1, 2 * k, y[i] + k + 1);
      imos(x[i] + 1, y[i] + k + 1, x[i] + k + 1, 2 * k);
    }
  }
  for (int i = 0; i <= 2 * k; i++) {
    for (int j = 1; j <= 2 * k; j++) {
      score[i][j] += score[i][j - 1];
    }
  } 
  for (int i = 0; i <= 2 * k; i++) {
    for (int j = 1; j <= 2 * k; j++) {
      score[j][i] += score[j - 1][i];
    }
  }
  int ans = 0;
  for (int i = 0; i < 2 * k; i++) {
    for (int j = 0; j < 2 * k; j++) {
      ans = max(ans, score[i][j]);
    }
  }
  cout << ans << endl;
  return 0;
}