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: Accepted)

#include <stdio.h>
#include <algorithm>
 
using namespace std;
 
int n, k;
int ps[2][2002][2002];
 
int sum(int r1, int c1, int r2, int c2, int w) {
	if(r1>r2 || c1>c2) return 0;
	int ret=ps[w][r2][c2];
	if(r1>0) ret-=ps[w][r1-1][c2];
	if(c1>0) ret-=ps[w][r2][c1-1];
	if(r1>0 && c1>0) ret+=ps[w][r1-1][c1-1];
	return ret;
}
 
int main() {
	scanf("%d %d",&n,&k);
	while(n--) {
		int x, y;
		char c;
		scanf("%d %d %c",&x,&y,&c);
		x%=2*k; y%=2*k;
		if(c=='W') ps[0][x][y]++;
		else ps[1][x][y]++;
	}
	for(int i=0;i<2*k;i++) {
		for(int j=0;j<2*k;j++) {
			for(int t=0;t<2;t++) {
				if(i>0) ps[t][i][j]+=ps[t][i-1][j];
				if(j>0) ps[t][i][j]+=ps[t][i][j-1];
				if(i>0 && j>0) ps[t][i][j]-=ps[t][i-1][j-1];
			}
		}
	}
	int res=0;
	for(int w=0;w<2;w++) {
		int d=sum(0,0,k-1,k-1,w)+sum(k,k,2*k-1,2*k-1,w)+
			sum(0,k,k-1,2*k-1,w^1)+sum(k,0,2*k-1,k-1,w^1);
		if(d>res) res=d;
	}
	for(int j=0;j+k+1<2*k;j++) {
		int r[3]={0,k,2*k}, c[4]={0,j+1,j+k+1,2*k};
		int sum1=0, sum2=0;
		for(int r1=0;r1<2;r1++) {
			for(int c1=0;c1<3;c1++) {
				if((r1+c1)%2) {
					sum1+=sum(r[r1],c[c1],r[r1+1]-1,c[c1+1]-1,0);
					sum2+=sum(r[r1],c[c1],r[r1+1]-1,c[c1+1]-1,1);
				} else {
					sum1+=sum(r[r1],c[c1],r[r1+1]-1,c[c1+1]-1,1);
					sum2+=sum(r[r1],c[c1],r[r1+1]-1,c[c1+1]-1,0);
				}
			}
		}
		if(max(sum1,sum2)>res) res=max(sum1,sum2);
	}
	for(int i=0;i+k+1<2*k;i++) {
		int r[4]={0,i+1,i+k+1,2*k}, c[3]={0,k,2*k};
		int sum1=0, sum2=0;
		for(int r1=0;r1<3;r1++) {
			for(int c1=0;c1<2;c1++) {
				if((r1+c1)%2) {
					sum1+=sum(r[r1],c[c1],r[r1+1]-1,c[c1+1]-1,0);
					sum2+=sum(r[r1],c[c1],r[r1+1]-1,c[c1+1]-1,1);
				} else {
					sum1+=sum(r[r1],c[c1],r[r1+1]-1,c[c1+1]-1,1);
					sum2+=sum(r[r1],c[c1],r[r1+1]-1,c[c1+1]-1,0);
				}
			}
		}
		if(max(sum1,sum2)>res) res=max(sum1,sum2);
	}
	for(int i=0;i+k+1<2*k;i++) {
		for(int j=0;j+k+1<2*k;j++) {
			int sum1=0, sum2=0;
			int r[4]={0,i+1,i+k+1,2*k}, c[4]={0,j+1,j+k+1,2*k};
			for(int r1=0;r1<3;r1++) {
				for(int c1=0;c1<3;c1++) {
					if((r1+c1)%2) {
						sum1+=sum(r[r1],c[c1],r[r1+1]-1,c[c1+1]-1,0);
						sum2+=sum(r[r1],c[c1],r[r1+1]-1,c[c1+1]-1,1);
					} else {
						sum1+=sum(r[r1],c[c1],r[r1+1]-1,c[c1+1]-1,1);
						sum2+=sum(r[r1],c[c1],r[r1+1]-1,c[c1+1]-1,0);
					}
				}
			}
			if(max(sum1,sum2)>res) res=max(sum1,sum2);
		}
	}
	printf("%d\n",res);
	return 0;
}