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>
#define int long long

using namespace std;
using LL = long long;
using P = pair<int, int>;

#define FOR(i, a, n) for(int i = (int)(a); i < (int)(n); ++i)
#define REP(i, n) FOR(i, 0, n)

#define pb(a) push_back(a)
#define all(x) (x).begin(),(x).end()

const int INF = (int)1e9;
const LL INFL = (LL)1e15;
const int MOD = 1e9 + 7;

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

/*************** using variables **************/
int n, k;
vector<int> x, y;
vector<char> c;
/**********************************************/

signed main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    cin >> n >> k;
    x.resize(n);
    y.resize(n);
    c.resize(n);
    REP(i, n) cin >> x[i] >> y[i] >> c[i];
    REP(i, n) if(c[i] == 'W') x[i] += k;
    REP(i, n) {
        x[i] %= 2*k;
        y[i] %= 2*k;
    }

    vector<vector<int>> imos(2*k + 5, vector<int>(2*k + 5, 0));
    REP(i, n){
        imos[x[i]+1][y[i]+1]++;
    }

    FOR(i, 1, 2*k+1) FOR(j, 1, 2*k+1) imos[i][j] += imos[i-1][j];
    FOR(i, 1, 2*k+1) FOR(j, 1, 2*k+1) imos[i][j] += imos[i][j-1];
    
    int ans = -INF;
    for(int i = 0; i <= k; i++){
        for(int j = 0; j <= k; j++){
            int a = imos[i][j];
            int b = imos[i+k][j+k] - imos[i][j+k] - imos[i+k][j] + imos[i][j];
            int c = imos[2*k][j] - imos[i+k][j];
            int d = imos[i][2*k] - imos[i][j+k];
            int e = imos[2*k][2*k] - imos[2*k][j+k] - imos[i+k][2*k] + imos[i+k][j+k];
            ans = max(ans, a+b+c+d+e);
        }
    }
    
    cout << ans << endl;
}