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<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct data{
	int x,y,v;
}a[210001];
char ch[11];
int i,j,k,n,m,w[1101][1101],t1,t2,ans;
void add(int x1,int y1,int x2,int y2,int v){
	x1++;y1++;x2++;y2++;
	w[x1][y1]++;w[x1][y2+1]--;w[x2+1][y1]--;w[x2+1][y2+1]++;
}
int main(){
	scanf("%d%d",&n,&m);memset(w,0,sizeof(w));
	for (i=1;i<=n;i++){
		scanf("%d%d%s",&a[i].x,&a[i].y,ch);a[i].v=(ch[0]=='B')?1:0;
		k=(a[i].x/m+a[i].y/m)%2;a[i].v^=k;
		a[i].x%=m;a[i].y%=m;
	}
	for (i=1;i<=n;i++){
		if (a[i].v){
			if (a[i].x<m-1) add(a[i].x+1,0,m-1,a[i].y,1);
			if (a[i].y<m-1) add(0,a[i].y+1,a[i].x,m-1,1);
		}else{
			add(0,0,a[i].x,a[i].y,1);
			if (a[i].x<m-1&&a[i].y<m-1) add(a[i].x+1,a[i].y+1,m-1,m-1,1);
		}
	}
	for (t1=2*n,t2=-2*n,i=1;i<=m;i++)
	 for (j=1;j<=m;j++){
	 	w[i][j]+=w[i-1][j]+w[i][j-1]-w[i-1][j-1];
	 	t1=min(t1,w[i][j]);t2=max(t2,w[i][j]);
	 }
	ans=max(t2,n-t1);
	printf("%d\n",ans);
	return 0;
}