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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int main(void) {
  int N, K;
  int **KW, **KB, **numW, **numB;
  scanf("%d %d", &N, &K);
  KW = (int**)calloc(sizeof(int*), (K + 1));
  KB = (int**)calloc(sizeof(int*), (K + 1));
  numW = (int**)calloc(sizeof(int*), (K + 1));
  numB = (int**)calloc(sizeof(int*), (K + 1));
  for (register int i = 0; i < (K + 1); i++) {
    KW[i] = (int*)calloc(sizeof(int), (K + 1));
    KB[i] = (int*)calloc(sizeof(int), (K + 1));
    numW[i] = (int*)calloc(sizeof(int), (K + 1));
    numB[i] = (int*)calloc(sizeof(int), (K + 1));
  }
  for (register int i = 0; i < N; i++) {
    int x, y;
    char c;
    scanf("%d %d %c", &x, &y, &c);
    if (c == 'W') {
      KW[y%K + 1][x%K + 1]++;
    } else {
      KB[y%K + 1][x%K + 1]++;
    }
  }

  for (int j = 1; j <= K; j++) {
    for (register int i = 1; i <= K; i++) {
      numW[j][i] = numW[j-1][i] + numW[j][i-1] + KW[j][i] - numW[j-1][i-1];
      numB[j][i] = numB[j-1][i] + numB[j][i-1] + KB[j][i] - numB[j-1][i-1];
    }
  }

  int max = 0;
  for (int j = 0; j <= K; j++) {
    for (int i = 0; i <= K; i++) {
      int cnt = numW[K][K] - numW[j][K] - numW[K][i] + 2 * numW[j][i]
        + numB[j][K] + numB[K][i] - 2 * numB[j][i];
      if (max < cnt) max = cnt;
    }
  }

  printf("%d\n", max);
  return 0;
}