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 <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <deque>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <utility>
#include <vector>

#define CK(N, A, B) (A <= N && N < B)
#define REP(i, a, b) for (int i = a; i < b; i++)
#define RREP(i, a, b) for (int i = (b - 1); a <= i; i--)
#define F first
#define S second
#define ll long long

const int INF = 1e9;
const long long LLINF = 1e18;

using namespace std;

int dy[] = {0,1,0,-1};
int dx[] = {1,0,-1,0};
int dy8[] = {0,1,1,1,0,-1,-1,-1};
int dx8[] = {1,1,0,-1,-1,-1,0,1};

/*==================================================*/
/*variable*/
int N, K;
int x, y;
char c;

int d[2010][2010];

/*==================================================*/
/*function*/
int cal(int x1, int y1, int x2, int y2){
    int ret = 0;

    if(x1 >= 2*K || y1 >= 2*K || x2 < 0 || y2 < 0) return ret;  //全くの範囲外

    if(x1 < 0) x1 = 0;  //はみ出しをカット
    if(y1 < 0) y1 = 0;
    if(x2 >= 2*K) x2 = 2*K-1;
    if(y2 >= 2*K) y2 = 2*K-1;

    ret = d[y2][x2];    //累積和で計算
    if(y1-1 >= 0) ret -= d[y1-1][x2];
    if(x1-1 >= 0) ret -= d[y2][x1-1];
    if(y1-1 >= 0 && x1-1 >= 0) ret += d[y1-1][x1-1];

    return ret;
}

/*==================================================*/

int main() {
    cin>>N>>K;
    REP(i,0,N){
        cin>>x>>y>>c;
        x %= (2*K);
        y %= (2*K);
        if(c == 'W'){
            y = (y + K) % (2*K);
        }
        d[y][x]++;
    }

    //２次元累積和の準備
    REP(i,0,2*K+1){
        REP(j,0,2*K+1){
            d[i][j+1] += d[i][j];
        }
    }
    REP(j,0,2*K){
        REP(i,0,2*K){
            d[i+1][j] += d[i][j];
        }
    }

    int ans = 0;

    //塗り方を全通り試す
    REP(i,0,2*K){
        REP(j,0,2*K){
            //各塗り方で黒い部分に含まれる点を数える（二次元累積和）
            int t = cal(i,j,i+K-1, j+K-1) +
                    cal(i-K,j-K,i-1, j-1) +
                    cal(i-K,j+K,i-1, j+2*K-1) +
                    cal(i+K,j-K,i+2*K-1, j-1) +
                    cal(i+K,j+K,i+2*K-1, j+2*K-1);
            ans = max(ans, t);
        }
    }

    cout<<ans<<endl;

    return 0;
}