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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse2")
#define ll          long long
#define pb          push_back
#define pii         pair<int,int>
#define vi          vector<int>
#define vii         vector<pii>
#define mi          map<int,int>
#define mii         map<int,pii>    
#define all(a)      (a).begin(),(a).end()
#define x           first
#define y           second
#define sz(x)       (int)x.size()
#define hell        1000000007
#define endl        '\n'
#define rep(i,a,b)  for(int i=a;i<b;i++)
using namespace std;
int v[100005];
vi x(100001),y(100001);
vi X[2001],Y[2001];
char c[100001];
void solve()
{
    int n,k,ans=0,cnt=0;
    cin>>n>>k;
    rep(i,0,n){
        cin>>x[i]>>y[i]>>c[i];
        X[x[i]%k].pb(i);
        Y[y[i]%k].pb(i);
        if((x[i]/k+y[i]/k)%2==0) v[i]=1;
        if(v[i] && c[i]=='B') cnt++;
        if(v[i]==0 && c[i]=='W') cnt++;
    }
    ans=cnt;
    int K=k+2;
    rep(a,0,K){
        rep(b,0,K){
            int f=b%k;
            for(auto i:Y[f]){
                v[i]=!v[i];
                if(v[i] && c[i]=='B') cnt++;
                else if(v[i]==0 && c[i]=='W') cnt++;
                else cnt--;
            }
            ans=max(ans,cnt);
        }
        int f=a%k;
        for(auto i:X[f]){
            v[i]=!v[i];
            if(v[i] && c[i]=='B') cnt++;
            else if(v[i]==0 && c[i]=='W') cnt++;
            else cnt--;
        }
        ans=max(ans,cnt);
    }
    cout<<ans;
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t=1;
    // cin>>t;
    while(t--)
        solve();
    return 0;
}