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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int, int> ii;
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define sqr(x) ((x) * (x))
#define X first
#define Y second
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define REP(i, n) FOR (i, 0, n)
#define TRACE(x) cout << #x << " = " << x << endl
#define _ << " _ " <<
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define MAXK 1010

int n, k;
int b[2 * MAXK][2 * MAXK], w[2 * MAXK][2 * MAXK], pb[2 * MAXK][2 * MAXK], pw[2 * MAXK][2 * MAXK];
int cntb, cntw;

int main()
{
  scanf("%d%d", &n, &k);
  int K = 2 * k;
  for (int i = 0; i < n; i++) {
    int x1, y1; char c1;
    scanf("%d%d %c", &x1, &y1, &c1);
    if (c1 == 'B') {
      b[x1 % K][y1 % K]++;
      cntb++;
    }
    else {
      w[x1 % K][y1 % K]++;
      cntw++;
    }
  }
  pb[0][0] = b[0][0];
  pw[0][0] = w[0][0];
  for (int i = 1; i < K; i++) {
    pb[0][i] = pb[0][i - 1] + b[0][i];
    pw[0][i] = pw[0][i - 1] + w[0][i];
    pb[i][0] = pb[i - 1][0] + b[i][0];
    pw[i][0] = pw[i - 1][0] + w[i][0];
  }
  for (int i = 1; i < K; i++) {
    for (int j = 1; j < K; j++) {
      pb[i][j] = pb[i - 1][j] + pb[i][j - 1] - pb[i - 1][j - 1] + b[i][j];
      pw[i][j] = pw[i - 1][j] + pw[i][j - 1] - pw[i - 1][j - 1] + w[i][j];
    }
  }
  /// WHITE FIX
  int rj = 0;
  for (int u = 1, d = k; d < K; u++, d++)
    for (int l = 1, r = k; r < K; r++, l++) {
      int wonw = pw[d][r] - pw[d][l - 1] - pw[u - 1][r] + pw[u - 1][l - 1] + pw[u - 1][l - 1] + pw[K - 1][l - 1] - pw[d][l - 1] + pw[u - 1][K - 1] - pw[u - 1][r] + pw[K - 1][K - 1] - pw[K - 1][r] - pw[d][K - 1] + pw[d][r];

      int bonw = pb[d][r] - pb[d][l - 1] - pb[u - 1][r] + pb[u - 1][l - 1] + pb[u - 1][l - 1] + pb[K - 1][l - 1] - pb[d][l - 1] + pb[u - 1][K - 1] - pb[u - 1][r] + pb[K - 1][K - 1] - pb[K - 1][r] - pb[d][K - 1] + pb[d][r];

      int bonb = cntb - bonw;
      int wonb = cntw - wonw;

      rj = max(rj, wonw + bonb);
    }
  /// BLACK FIX
  for (int u = 1, d = k; d < K; u++, d++)
    for (int l = 1, r = k; r < K; r++, l++) {
      int wonb = pw[d][r] - pw[d][l - 1] - pw[u - 1][r] + pw[u - 1][l - 1] + pw[u - 1][l - 1] + pw[K - 1][l - 1] - pw[d][l - 1] + pw[u - 1][K - 1] - pw[u - 1][r] + pw[K - 1][K - 1] - pw[K - 1][r] - pw[d][K - 1] + pw[d][r];

      int bonb = pb[d][r] - pb[d][l - 1] - pb[u - 1][r] + pb[u - 1][l - 1] + pb[u - 1][l - 1] + pb[K - 1][l - 1] - pb[d][l - 1] + pb[u - 1][K - 1] - pb[u - 1][r] + pb[K - 1][K - 1] - pb[K - 1][r] - pb[d][K - 1] + pb[d][r];

      int wonw = cntw - wonb;
      int bonw = cntb - bonb;

      rj = max(rj, wonw + bonb);
    }
  cout << rj;
  return 0;
}