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 abs(int a) {
  if (a >= 0) {
    return a;
  } else {
    return (-1)*a;
  }
}

int even(int a) {
  if (a%2 == 0)
    return 0; //偶数0
  else
    return 1; //奇数1
}

int main(void)
{
  
  //入力
  int N, K;
  scanf("%d %d", &N, &K);
  int x[N], y[N];
  char c[N];
  for (int i = 0; i < N; i++) {
    scanf("%d %d %s", &x[i], &y[i], &c[i]);
  }
  
  //計算
  //ホワイト
  int cntW = 0;
  for (int i = (-1)*K+1; i < K; i++) {
    for (int j = (-1)*K+1; j < K; j++) {
      int cnt = 0;
      for (int k = 0; k < N; k++) {
        if (c[k] == 'W' && even((x[k]-i)/K) == 0 && even((y[k]-j)/K) == 0) cnt++;
        if (c[k] == 'B' && even((x[k]-i)/K) == 1 && even((y[k]-j)/K) == 1) cnt++;
        if (cnt > cntW ) cntW = cnt;
      }
    }
  }

  //ブラック
  int cntB = 0;
  for (int i = (-1)*K+1; i < K; i++) {
    for (int j = (-1)*K+1; j < K; j++) {
      int cnt = 0;
      for (int k = 0; k < N; k++) {
        if (c[k] == 'B' && even((x[k]-i)/K) == 0 && even((y[k]-j)/K) == 0) cnt++;
        if (c[k] == 'W' && even((x[k]-i)/K) == 1 && even((y[k]-j)/K) == 1) cnt++;
        if (cnt > cntB ) cntB  = cnt;
      }
    }
  }
  
  int max = 0;
  if (cntW > cntB) max = cntW;
  else max = cntB;
  
  printf("%d\n", max);

  return 0;
  
}