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;
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define int long long
int n,k,x[100007],y[100007],id[100007],fl[2007][2007],sum[2007][2007],ans=0;
char c[100007];
bool cmp(int a,int b){
	return (x[a]<x[b] || (x[a]==x[b] && y[a]<y[b]) || (x[a]==x[b] && y[a]==y[b] && a<b));
}
int get(int lx,int ly,int rx,int ry){
	if(lx>rx || ly>ry) return 0;
	return sum[rx][ry]-sum[rx][ly-1]-sum[lx-1][ry]+sum[lx-1][ly-1];
}
const int inf = 1e9;
signed main(){
	ios_base::sync_with_stdio(0);
	cin>>n>>k;
	for(int i=1;i<=n;i++){
		cin>>x[i]>>y[i]>>c[i];
		if(c[i]=='B') x[i]-=k;
		id[i]=i;
	}
	sort(id+1,id+1+n,cmp);
	int sx=x[id[1]],sy=y[id[1]];
	fl[1][1]++;
	for(int j=2;j<=n;j++){
		int i=id[j];
		int cx=(x[i]-sx)%(2*k);
		int cy=(y[i]-sy+(2*k*inf))%(2*k);
		if(cx<0) cx+=2*k;
		if(cy<0) cy+=2*k;
		cx++;
		cy++;
		fl[cx][cy]++;
		//cout<<((x[i]-sx)%(2*k))+1<<" "<<((y[i]-sy)%(2*k))+1<<endl;
	}
	ans=0;
	memset(sum,0,sizeof(sum));
	sum[1][1]=fl[1][1];
	for(int i=2;i<=2*k;i++){
		sum[i][1]=sum[i-1][1]+fl[i][1];
		sum[1][i]=sum[1][i-1]+fl[1][i];
	}
	for(int i=2;i<=2*k;i++){
		for(int j=2;j<=2*k;j++){
			sum[i][j]=sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1]+fl[i][j];
		}
	}
	for(int i=1;i<=k;i++){
		for(int j=1;j<=k;j++){
			int val=get(1,1,i,j)+get(i+k+1,1,2*k,j)+get(1,j+k+1,i,2*k)+get(i+1,j+1,i+k,j+k);
			ans=max(ans,max(val,n-val));
			//cout<<val<<" ";
		}
		//cout<<endl;
	}
	cout<<ans;
}