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;
int lt[2050][2050];
int  xx[10050];
int  yy[10050];
char c0[10050];
int main () {
	int n,k,x,y;
	char c;
	cin>>n>>k;
	if(n>8000||k>1500) {
		int  xx;
		int  yy;
		for(int i=0; i<n; i++) {
			scanf("%d %d %c",&xx,&yy,&c);
			if(c=='W') {
				yy+=k;
			}
			xx%=2*k;
			yy%=2*k;
			lt[xx][yy]++;
		}
		for(int i=0; i<=2*k; i++)
			for(int j=1; j<=2*k; j++)
				lt[i][j]+=lt[i][j-1];
		for(int i=1; i<=2*k; i++)
			for(int j=0; j<=2*k; j++)
				lt[i][j]+=lt[i-1][j];
		int  ans=0;
		for(int a=0; a<=k; a++) {
			for(int b=0; b<=k ; b++) {
				int c1=lt[b][a];
				int c2=lt[b][2*k]-lt[b][k+a];
				int c3=lt[2*k][a]-lt[k+b][a];
				int c4=lt[2*k][2*k]-lt[k+b][2*k]-lt[2*k][k+a]+lt[k+b][a+k];
				int c5=lt[k+b][a+k]-lt[b][a+k]-lt[b+k][a]+lt[b][a];
				ans=max(ans,c1+c2+c3+c4+c5);
				ans=max(ans,n-(c1+c2+c3+c4+c5));
			}
		}
		printf("%d\n",ans);
	} else {
		for(int i=0; i<n; i++) {
			scanf("%d %d %c",&xx[i],&yy[i],&c0[i]);
			xx[i]+=k;
			yy[i]+=k;
		}
		int  ans=0,ct,ct2;
		for(int sx=0; sx<k; sx++) {
			for(int sy=0; sy<k; sy++) {
				ct=0,ct2=0;
				for(int i=0; i<n; i++) {
					int dx=abs(xx[i]-sx)%(2*k);
					int dy=abs(yy[i]-sy)%(2*k);
					if((c0[i]=='W'&&((dx<k&&dy<k)||(dx>=k&&dy>=k)))||(c0[i]=='B'&&((dx>=k&&dy<k)||(dx<k&&dy>=k)))) {
						ct++;
					}
					if((c0[i]=='B'&&((dx<k&&dy<k)||(dx>=k&&dy>=k)))||(c0[i]=='W'&&((dx>=k&&dy<k)||(dx<k&&dy>=k)))) {
						ct2++;
					}
				}
				ans=max(ans,max(ct,ct2));
			}
		}
		printf("%d\n",ans);
		return 0;
	}
	return 0;
}