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 <iostream>
#include <fstream>
#include <iomanip>
#include <math.h>
#include <limits.h>
#include <algorithm>
#include <vector>
#include <list>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
#include <string>
#include <string.h>
#include <sstream>
#include <ctime>

using namespace std;

#define eps 1e-12
#define pi 3.14159265358979323846
#define pb push_back
#define mp make_pair
#define st first
#define nd second
#define bgn begin
#define ll long long
#define ld long double
#define ull unsigned long long
#define ii pair<ll,ll>









const int N=2010;
int n,k,k2,s[N][N],x,y,dx,dy,res;
char c,c2;


void f(int j,int inc)
{
	s[0][j]+=inc;
	s[dx+1][j]-=inc;
	s[dx+k+1][j]+=inc;
}

void f2(int j,int inc)
{
	s[dx+1][j]+=inc;
	s[dx+k+1][j]-=inc;
}




void solve()
{
	cin>>n>>k;
	k2=2*k;
	for(int i=0;i<=k2+1;i++)for(int j=0;j<=k2+1;j++)s[i][j]=0;
	for(int i=1;i<=n;i++)
	{
		cin>>x>>y>>c;
		dx=x%k2;
		dy=y%k2;
		if(dx<k)
		{
			if(dy<k)c2='W';
			else c2='B';
		}
		else
		{
			if(dy<k)c2='B';
			else c2='W';
		}
		if(dy>=k)dy-=k;
		if(dx>=k)dx-=k;
		if(c==c2)
		{
			f(0,1);
			f(dy+1,-1);
			f(dy+k+1,1);
			f2(dy+1,1);
			f2(dy+k+1,-1);
		}
		else
		{
			f2(0,1);
			f2(dy+1,-1);
			f2(dy+k+1,1);
			f(dy+1,1);
			f(dy+k+1,-1);
		}
	}
	for(int j=0;j<k2;j++)
	{
		for(int i=1;i<k2;i++)s[i][j]+=s[i-1][j];
	}
	res=0;
	for(int i=0;i<k2;i++)
	{
		res=max(res,s[i][0]);
		for(int j=1;j<k2;j++)
		{
			s[i][j]+=s[i][j-1];
			res=max(res,s[i][j]);
		}
	}
	cout<<res<<"\n";
}

int main()
{
	std::ios::sync_with_stdio(0);
	cin.tie(0);
#ifdef localProject
	freopen("in.txt","r",stdin);
#endif
	solve();
	return 0;
}