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>
#include <iomanip>
using namespace std;
#define reps(i,s,n) for(int i = s; i < n; i++)
#define rep(i,n) reps(i,0,n)
#define fi first
#define se second
#define mp make_pair
typedef long long ll;
typedef vector<ll> vec;
typedef vector<vec> mat;

ll N,M,H,W,K,Q,A,B,C,L,R;
string S,T;
const ll MOD = (1e+9) + 7;
const ll INF = 1LL << 60;
typedef pair<ll,ll> P;

typedef vector<P> vp;
typedef vector<vp> matP;


int main(){
    cin>>N>>K;
    vector<mat> memo(2,mat(K, vec(K,0))), sum(2,mat(K+1, vec(K+1,0)));
    ll K2 = K*2;
    char c;
    rep(i,N){
        cin>>A>>B>>c;
        ll white = (c == 'W');
        A %= K2; B %= K2;
        (white += A/K + B/K)%=2;
        ++memo[white][A%K][B%K];
    }
    rep(i,2){
        rep(j,K){
            rep(k,K){
                ll temp = memo[i][j][k];
                temp += sum[i][j+1][k];
                temp += sum[i][j][k+1];
                temp -= sum[i][j][k];
                sum[i][j+1][k+1] = temp;
            }
        }
    }
    rep(i,2){
        rep(k,K+1){
            rep(j,K+1) cout<<sum[i][j][k]<<' ';
            cout<<endl;
        }
        cout<<endl;
    }
    ll ans = 0;
    rep(i,K+1){
        rep(j,K+1){
            ll temp = sum[0][i][j];
            temp += sum[1][i][K] - sum[1][i][j];
            temp += sum[1][K][j] - sum[1][i][j];
            temp += sum[0][K][K] - sum[0][i][K] - sum[0][K][j] + sum[0][i][j];
            ans = max(ans, temp);
        }
    }
    rep(i,K+1){
        rep(j,K+1){
            ll temp = sum[1][i][j];
            temp += sum[0][i][K] - sum[0][i][j];
            temp += sum[0][K][j] - sum[0][i][j];
            temp += sum[1][K][K] - sum[1][i][K] - sum[1][K][j] + sum[1][i][j];
            ans = max(ans, temp);
        }
    }
    cout<<ans<<endl;
}