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>
#include<stdlib.h>
#include<stdint.h>
#include<inttypes.h>

typedef int32_t i32;

#define POS(i, j) ((i) * (2 * k + 1) + (j))

//[a,b) * [c, d)
i32 calc (i32 *sum, i32 k, i32 a, i32 b, i32 c, i32 d) {
  return sum[POS(b, d)] - sum[POS(b, c)] - sum[POS(a, d)] + sum[POS(a, c)];
}

void run (void) {
  i32 n, k;
  scanf ("%" SCNi32 "%" SCNi32, &n, &k);
  i32 *cnt = (i32 *) calloc ((2 * k + 1) * (2 * k + 1), sizeof (i32));
  for (i32 i = 0; i < n; ++i) {
    i32 x, y;
    char s[5];
    scanf ("%" SCNi32 "%" SCNi32 "%s", &x, &y, s);
    if (s[0] == 'B') x += k;
    cnt[POS(x % (2 * k), y % (2 * k))]++;
  }
  i32 *sum = (i32 *) calloc ((2 * k + 1) * (2 * k + 1), sizeof (i32));
  for (i32 i = 1; i <= 2 * k; ++i) {
    for (i32 j = 1; j <= 2 * k; ++j) {
      sum[POS(i, j)] = sum[POS(i - 1, j)] + sum[POS(i, j - 1)] - sum[POS(i - 1, j - 1)] + cnt[POS(i - 1, j - 1)];
    }
  }
  i32 max = 0;
  for (i32 i = 0; i <= k; ++i) {
    for (i32 j = 0; j <= k; ++j) {
      i32 a = calc (sum, k, 0    , i    , 0    , j    );
      i32 b = calc (sum, k, i    , i + k, j    , j + k);
      i32 c = calc (sum, k, i + k, 2 * k, 0    , j    );
      i32 d = calc (sum, k, 0    , i    , j + k, 2 * k);
      i32 e = calc (sum, k, i + k, 2 * k, j + k, 2 * k);
      i32 q = a + b + c + d + e;
      if (max < q) max = q;
    }
  }
  printf ("%" PRIi32 "\n", max);
}

int main (void) {
  run();
  return 0;
}