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<algorithm>
using namespace std;

inline int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0' || '9'<ch){if(ch=='-')f=-1;ch=getchar();}
	while('0'<=ch && ch<='9')x=x*10+(ch^48),ch=getchar();
	return x*f;
}

inline void writes(int x)
{
	if(x>=10)writes(x/10);
	putchar('0'+x%10);
}

inline void write(int x,char ch='\n')
{
	if(x<0)putchar('-'),writes(-x);
	else writes(x);putchar(ch);
}

const int N=1e5+9;
const int K=2009;
const int KK=(K<<1)+9;

int n,k,kk;
int x[N],y[N];
int w[KK][KK],b[KK][KK];

inline int query(int (*a)[KK],int x1,int y1,int x2,int y2)
{
	x2=min(x2,kk),y2=min(y2,kk);
	int ret=0;
	if(x2>=0 && y2>=0)ret=a[x2][y2];
	if(x1>0 && y2>=0)ret-=a[x1-1][y2];
	if(y1>0 && x2>=0)ret-=a[x2][y1-1];
	if(x1>0 && y1>0)ret+=a[x1-1][y1-1];
	return ret;
}

int main()
{
	if(fopen("vj.in","r"))
	{
		freopen("vj.in","r",stdin);
		freopen("vj.out","w",stdout);
	}

	n=read();k=read();
	char c[190];
	kk=k*2;
	for(int i=1;i<=n;i++)
	{
		x[i]=read();y[i]=read();scanf("%s",c+1);
		if(c[1]=='B')
			b[x[i]%kk][y[i]%kk]++;
		else
			w[x[i]%kk][y[i]%kk]++;
	}

	for(int i=0,j=1;j<=kk;j++)
	{
		b[i][j]+=b[i][j-1];
		w[i][j]+=w[i][j-1];
	}
	for(int i=1,j=0;i<=kk;i++)
	{
		b[i][j]+=b[i-1][j];
		w[i][j]+=w[i-1][j];
	}
	for(int i=1;i<=kk;i++)
		for(int j=1;j<=kk;j++)
		{
			b[i][j]+=b[i-1][j]+b[i][j-1]-b[i-1][j-1];
			w[i][j]+=w[i-1][j]+w[i][j-1]-w[i-1][j-1];
		}

	/*
	for(int i=0;i<=6;i++,puts(""))
		for(int j=0;j<=6;j++)
			printf("%d ",b[i][j]);
	puts("");
	for(int i=0;i<=6;i++,puts(""))
		for(int j=0;j<=6;j++)
			printf("%d ",w[i][j]);
	*/

	int ans=0;
	for(int tim=0;tim<=1;tim++)
	{
	for(int i=0;i<=k;i++)
		for(int j=0;j<=k;j++)
		{
			int satw=query(w,i,j,i+k-1,j+k-1)+query(w,i+k,j+k,i+k+k-1,j+k+k-1);
			satw+=query(w,i+k,0,i+k+k-1,j-1)+query(w,0,j+k,i-1,j+k+k-1);
			if(i>0 && j>0)satw+=w[i-1][j-1];

			int satb=query(b,i+k,j,i+k+k-1,j+k-1)+query(b,i,j+k,i+k-1,j+k+k-1);
			satb+=query(b,i,0,i+k-1,j-1)+query(b,0,j,i-1,j+k-1);

			ans=max(ans,satw+satb);
		}

	swap(w,b);
	}

	
	printf("%d\n",ans);
	return 0;
}