Problem: 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:

cba927b2484fad94fb5ff7473e9aadef.png

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.

Examples

Input

4 3
0 1 W
1 2 W
5 3 B
5 4 B


Output

4


Input

2 1000
0 0 B
0 1 W


Output

2


Input

6 2
1 2 B
2 1 W
2 2 B
1 0 B
0 6 W
4 5 W


Output

4

Solution (in CPP):

Solution:
#include<set>
#include<cstdio>
#include<queue>
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
long long read(){
	long long f=1,x=0;
	char c=getchar();
	while(c<'0'||'9'<c){if(c=='-')f=-1;c=getchar();}
	while('0'<=c&&c<='9'){x=x*10+c-'0';c=getchar();}
	return f*x;
}
#define MAXN 100000
struct node{
	int x,y;
	bool color;//0 White 1 Black
}wish[MAXN+5];
int fit[MAXN+5],unfit[MAXN+5];
int main(){
	char c;
	int n=read(),k=read(),tmp,ans=0;
	for(int i=1;i<=n;i++)
	{
		wish[i].x=read()%(2*k),wish[i].y=read()%(2*k),c=getchar();
		if(c=='W')
			wish[i].color=1;
		else
			wish[i].color=0;
	}
	for(int x=0;x<=k;x++)
	{
		tmp=0;
		memset(fit,0,sizeof fit);
		memset(unfit,0,sizeof unfit);
		for(int i=1;i<=n;i++)
		{
			int p=(x<=wish[i].x&&wish[i].x<x+k),
				q=(0<=wish[i].y&&wish[i].y<k);
			if((p^q)==wish[i].color)
			{
				tmp++;
				fit[wish[i].y]++;
			}
			else
				unfit[wish[i].y]++;
		}
		for(int i=0;i<=k;i++)
		{
			if(tmp>ans) ans=tmp;
			if(n-tmp>ans) ans=n-tmp;
			if(i!=k)
				tmp=tmp-fit[i]+unfit[i]-fit[i+k]+unfit[i+k];		
		}
	}
	printf("%d\n",ans);
	return 0;
}