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) {
  long long int n, k, i, j, ans = 0, l, x, y;
  scanf("%lld%lld", &n, &k);
  long long int t[k][k][2], sum[k][k][2], sum2[k][k][2];
  for(i = 0; i < k; ++i) {for(j = 0; j < k; ++j) {for(l = 0; l < 2; ++l) t[i][j][l] = 0;}}
  char c;
  for(i = 0; i < n; ++i) {
    int s;
    scanf("%lld %lld %c", &x, &y, &c);
    if(c == 'B') s = 0;
    else s = 1;
    x %= 2 * k;
    y %= 2 * k;
    if(x >= k) {
      s = (s + 1) % 2;
      x -= k;
    }
    if(y >= k) {
      s = (s + 1) % 2;
      y -= k;
    }
    t[x][y][s]++;
  }
  for(i = 0; i < k; ++i) for(j = 0; j < 2; ++j) {sum[i][0][j] = t[i][0][j]; sum2[0][i][j]= t[0][i][j];}
  for(i = 0; i < k; ++i) for(j = 1; j < k; ++j) for(l = 0; l < 2; ++l) {sum[i][j][l] = sum[i][j - 1][l] + t[i][j][l]; sum2[j][i][l] = sum2[j - 1][i][l] + t[j][i][l];}
  /*for(i = 0; i < k; ++i) {
    for(j = 0; j < k; ++j) printf("(%d %d)", t[i][j][0], t[i][j][1]);
    printf("\n");
  }
  printf("\n");
  for(i = 0; i < k; ++i) {
    for(j = 0; j < k; ++j) printf("(%d %d)", sum2[i][j][0], sum2[i][j][1]);
    printf("\n");
  }*/
  for(i = 0; i <= k; ++i) {
    long long temp = 0;
    for(j = 0; j <= k; ++j){
       if(!j) {
       temp = 0;
       for(l = 0; l < k; ++l) {
         if(l < j) {
           if(i) {
           temp += sum[l][i - 1][0];
           temp += sum[l][k - 1][1] - sum[l][i - 1][1];
           } else {
           temp += sum[l][k - 1][1];
           }
         } else {
           if(i) {
           temp += sum[l][i - 1][1];
           temp += sum[l][k - 1][0] - sum[l][i - 1][0];
           } else {
           temp += sum[l][k - 1][0];
           }
         }
       }
       } else {
        if(!i) {
          temp -= sum[j - 1][k - 1][0];
          temp += sum[j - 1][k - 1][1];
        } else {
          temp -= sum[j - 1][i - 1][1] + sum[j - 1][k - 1][0] - sum[j - 1][i - 1][0];
          temp += sum[j - 1][i - 1][0] + sum[j - 1][k - 1][1] - sum[j - 1][i - 1][1];
        }
       }
       if(ans < temp) {
        ans = temp;
      } 
    }
  }
  printf("%lld", ans);
  return 0;
}