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 unsigned long long ull;
typedef long long ll;

int d[2000][2000];

int fcn(int x1, int y1, int x2, int y2){
    if(x1==0 && y1==0) return d[x2][y2];
    if(x1>0 && y1==0) return d[x2][y2]-d[x1-1][y2];
    if(x1==0 && y1>0) return d[x2][y2]-d[x2][y1-1];
    return d[x2][y2]-d[x2][y1-1]-d[x1-1][y2]+d[x1-1][y1-1];
}

int main(){
    int n, k;
    cin >> n >> k;
    ll x[n], y[n];
    for(int i=0; i<n; i++){
        char c;
        cin >> x[i] >> y[i] >> c;
        if(c=='W') x[i]+=k;
        x[i]%=2*k;
        y[i]%=2*k;
    }

    for(int i=0; i<n; i++){
        d[x[i]][y[i]]++;
    }
    for(int i=0; i<2*k; i++){
        for(int j=1; j<2*k; j++){
            d[i][j]+=d[i][j-1];
        }
    }
    for(int j=0; j<2*k; j++){
        for(int i=1; i<2*k; i++){
            d[i][j]+=d[i-1][j];
        }
    }

    int ans=0;
    for(int i=0; i<k; i++){
        for(int j=0; j<k; j++){
            int s=fcn(i, j, i+k-1, j+k-1)+fcn(i+k, j+k, 2*k-1, 2*k-1);
            if(i>0 && j>0) s+=fcn(0, 0 , i-1, j-1);
            if(i>0) s+=fcn(0, j+k, i-1, 2*k-1);
            if(j>0) s+=fcn(i+k, 0, 2*k-1, j-1);
            ans=max(ans, s);
        }
    }
    cout << ans << endl;
    return 0;
}