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;

#define DUMP(x) std::cerr << (#x) << " = " << (x) << "\n"
#define rep(i, n) for (int i = 0; i < (int)(n); ++i)
#define REP(i, k, n) for (int i = (k); i < (int)(n); ++i)
#define ALL(r) r.begin(), r.end()
#define YES puts("YES")
#define Yes puts("Yes")
#define NO puts("NO")
#define No puts("No")
#define IMP puts("IMPOSSIBLE")
#define Imp puts("Impossible")
#define imp puts("impossible")
#define M1 puts("-1")
#define pb push_back

template <typename T>
T dup(T x, T y)
{
    return (x + y - 1) / y;
};
template <typename A, size_t N, typename T>
inline void arrayFill(A (&array)[N], const T &val) { std::fill((T *)array, (T *)(array + N), val); }

using ll = int64_t;
using vint = vector<int32_t>;
using vvint = vector<vint>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vstr = vector<string>;
using pint = pair<int32_t, int32_t>;
using pll = pair<ll, ll>;
using setint = set<int32_t>;
using setstr = set<string>;
using qint = queue<int32_t>;
using qpint = queue<pint>;

constexpr std::int32_t INF = 1001001001;
constexpr std::int64_t LINF = 1001001001001001001;

constexpr int K = 1005;
vvint s(2 * K, vint(2 * K, 0));
vvint table(2 * K, vint(2 * K, 0));
// [x1, x2), [y1, y2)
int query(int x1, int y1, int x2, int y2)
{
    return s[x2][y2] - s[x1][y2] - s[x2][y1] + s[x1][y1];
}

void Main()
{
    int n, k;
    cin >> n >> k;
    rep(i, n)
    {
        int x, y;
        char c;
        cin >> x >> y >> c;
        if (c == 'W')
        {
            y += k;
        }
        x %= k * 2;
        y %= k * 2;
        ++table[x][y];
    }

    rep(x, 2 * k) rep(y, 2 * k)
    {
        s[x + 1][y + 1] = s[x][y + 1] + s[x + 1][y] - s[x][y] + table[x][y];
    }

    int ans = 0;
    rep(i, k + 1) rep(j, k + 1)
    {
        int ra = query(0, 0, i, j);
        int rb = query(i + k, 0, 2 * k, j);
        int rc = query(i, j, i + k, j + k);
        int rd = query(0, j + k, i, 2 * k);
        int re = query(i + k, j + k, 2 * k, 2 * k);
        ans = max(ra + rb + rc + rd + re, ans);
    }
    cout << ans << endl;
}

int main()
{
    std::cin.tie(nullptr);
    std::ios_base::sync_with_stdio(false);
    std::cout << std::fixed << std::setprecision(15);
    Main();
    return 0;
}