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: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;

typedef pair<int, int> II;
typedef vector<II> VII;
typedef vector<int> VI;
typedef vector<VI> VVI;

typedef long long int LL;

#define PB push_back
#define MP make_pair
#define F first
#define S second
#define SZ(a) (int)(a.size())
#define ALL(a) a.begin(), a.end()
#define SET(a, b) memset(a, b, sizeof(a))
#define si(n) scanf("%d", &n)
#define dout(n) printf("%d\n", n)
#define sll(n) scanf("%lld", &n)
#define lldout(n) printf("%lld\n", n)

#define TRACE

#ifdef TRACE
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1> void __f(const char *name, Arg1 &&arg1) {
  cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char *names, Arg1 &&arg1, Args &&... args) {
  const char *comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
#else
#define trace(...)
#endif

#define fast_io                                                                \
  ios_base::sync_with_stdio(false);                                            \
  cin.tie(NULL)

#define endl '\n'
const int N = 1e5 + 10;
const int K = 1e3 + 10;

int x[N], y[N], col[N], sat[N];

int rx[K][2], ry[K][2];

int main() {
  fast_io;
  int n, k;
  cin >> n >> k;
  int twok = 2 * k;
  int sat_now = 0;
  for (int i = 0; i < n; i++) {
    string co;
    cin >> x[i] >> y[i] >> co;
    if (co == "W")
      col[i] = 1;
    else
      col[i] = 0;
    int ccol = 0;
    if (x[i] % twok >= k)
      ccol ^= 1;
    if (y[i] % twok >= k)
      ccol ^= 1;
    if (ccol == col[i])
      sat[i] = 1;
    rx[x[i] % k][sat[i]]++;
    ry[y[i] % k][sat[i]]++;
    sat_now += sat[i];
  }
  int max_sat = sat_now;
  for (int sx = 0; sx < twok - 1; sx++) {
    int sat_now_b = sat_now;
    for (int sy = 0; sy < twok - 1; sy++) {
      int cy = (sy + 1) % k;
      sat_now += ry[cy][0] - ry[cy][1];
      max_sat = max(max_sat, sat_now);
    }
    sat_now = sat_now_b;
    int cx = (sx + 1) % k;
    sat_now += rx[cx][0] - rx[cx][1];
    max_sat = max(max_sat, sat_now);
  }
  cout << max_sat << endl;
  return 0;
}