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)

// header {{{
#include <bits/stdc++.h>
using namespace std;

// {U}{INT,LONG,LLONG}_{MAX,MIN}
#define INF         INT_MAX/3
#define LLINF       LLONG_MAX/3
#define MOD         (1000000007LL)
#define MODA(a, b)  a=((a)+(b))%MOD
#define MODP(a, b)  a=((a)*(b))%MOD
#define inc(i, l, r)   for(long long i=(l);i<(r);i++)
#define dec(i, l, r)   for(long long i=(r)-1;i>=(l);i--)
#define pb          push_back
#define se          second
#define fi          first
#define mset(a, b)  memset(a, b, sizeof(a))

using LL  = long long;
using G   = vector<vector<int>>;

int di[] = {0, -1, 0, 1};
int dj[] = {1, 0, -1, 0};
// }}}

int main() {
    std::ios::sync_with_stdio(false);
    int n, k;cin >> n >> k;
    vector<int> x(n), y(n);
    vector<char> c(n);
    inc(i, 0, n) cin >> x[i] >> y[i] >> c[i];
    map<pair<pair<int, int>, char>, int> st;
    inc(i, 0, n){
        int m = x[i]/k + y[i]/k;
        char cc = c[i];
        if(m%2 != 0){
            if(cc == 'B'){
                cc = 'W';
            }else{
                cc = 'B';
            }
        }
        st[{{x[i]%k, y[i]%k}, cc}]++;
    }
    //cout << endl;
    //for(auto kv : st){
        //cout << kv.fi.fi.fi << " " << kv.fi.fi.se << " " << kv.fi.se << kv.se << endl;
    //}
    int ans = 0;
    inc(i, 0, k+1){
        inc(j, 0, k+1){
            int cnt = 0;
            for(auto kv : st){
                auto s = kv.fi;
                int v = kv.se;
                int x = s.fi.fi;
                int y = s.fi.se;
                int c = s.se == 'B' ? 1 : 0;

                int d = 1-((x >= j) + (y >= i))%2;
                if(c == d) cnt+=v;
            }
            ans = max(ans, cnt);
        }
    }
    cout << ans << endl;
    return 0;
}