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;

#define REP(i,n) for(int i=0;i<n;++i)
#define SORT(name) sort(name.begin(), name.end())
#define ZERO(p) memset(p, 0, sizeof(p))
#define MINUS(p) memset(p, -1, sizeof(p))
#if 1
#  define DBG(fmt, ...) printf(fmt, __VA_ARGS__)
#else
#  define DBG(fmt, ...)
#endif

#define MOD 1000000007
#define INF 1000000000
#define MAX_N 100010
#define MAX_K 1010

// 二次元累積和
// コンストラクト時に幅と高さを与える必要がある
template< class T >
class CumulativeSum2D
{
public:
    vector< vector< T > > m_data;

    CumulativeSum2D(int W, int H) : m_data(W + 1, vector< int >(H + 1, 0)) {}

    // 要素 (x,y) に値 z を加える O(1)
    void Add(int x, int y, int z) {
        ++x, ++y;
        if(x >= m_data.size() || y >= m_data[0].size()) { return; }
        m_data[x][y] += z;
    }

    // 累積和の構築 O(WH)
    void Build() {
        for(int i = 1; i < m_data.size(); i++) {
            for(int j = 1; j < m_data[i].size(); j++) {
                m_data[i][j] += m_data[i][j - 1] + m_data[i - 1][j] - m_data[i - 1][j - 1];
            }
        }
    }

    // 左上 [sx, sy], 右下(gx, gy) の矩形内の和を求める O(1)
    // 右下は含まないので注意
    T Query(int sx, int sy, int gx, int gy) {
        return (m_data[gx][gy] - m_data[sx][gy] - m_data[gx][sy] + m_data[sx][sy]);
    }
};

int N, K;

int main()
{
    cin >> N >> K;
    CumulativeSum2D<int> cs_b(2 * K, 2 * K);   // 黒用
    CumulativeSum2D<int> cs_w(2 * K, 2 * K);   // 白用
    REP(i, N) {
        ll x, y;
        char c;
        cin >> x >> y >> c;
        if(c == 'W') { cs_w.Add((int)(x % (2 * K)), (int)(y % (2 * K)), 1); }
        else { cs_b.Add((int)(x % (2 * K)), (int)(y % (2 * K)), 1); }
    }
    cs_w.Build();
    cs_b.Build();
    // DEBUG
    /*
    printf("-- white --\n");
    REP(y, K) {
        REP(x, K) {
            DBG("%d ", cs_w.Query(x, y, x+1, y+1));
        }
        DBG("\n")
    }
    printf("-- black --\n");
    REP(y, K) {
        REP(x, K) {
            DBG("%d ", cs_b.Query(x, y, x+1, y+1));
        }
        DBG("\n")
    }
    */
    int ans = 0;
    REP(y, K) {
        REP(x, K) {
            int white1 = cs_w.Query(0, 0, x+1, y+1) + cs_w.Query(x+1+K, 0, 2*K, y+1)
                         + cs_w.Query(x+1, y+1, x+1+K, y+1+K) + cs_w.Query(0, y+1+K, x+1, 2*K)
                         + cs_w.Query(x+1+K, y+1+K, 2*K, 2*K);
            int black2 = cs_b.Query(0, 0, x+1, y+1) + cs_b.Query(x+1+K, 0, 2*K, y+1)
                         + cs_b.Query(x+1, y+1, x+1+K, y+1+K) + cs_b.Query(0, y+1+K, x+1, 2*K)
                         + cs_b.Query(x+1+K, y+1+K, 2*K, 2*K);
            int white2 = cs_w.Query(x+1, 0, x+1+K, y+1) + cs_w.Query(0, y+1, x+1, y+1+K)
                         + cs_w.Query(x+1+K, y+1, 2*K, y+1+K) + cs_w.Query(x+1, y+1+K, x+1+K, 2*K);
            int black1=  cs_b.Query(x+1, 0, x+1+K, y+1) + cs_b.Query(0, y+1, x+1, y+1+K)
                         + cs_b.Query(x+1+K, y+1, 2*K, y+1+K) + cs_b.Query(x+1, y+1+K, x+1+K, 2*K);

            ans = max( { ans, (white1 + black1), (white2 + black2) } );
        }
    }

    printf("%d\n", ans);
    return 0;
}