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<iostream>
#include<cstring>
#include<cstdio>
#define in(x) scanf("%d",&x)
#define ic(x) cin>>x
#define N 100010
#define K 1010
using namespace std;
int n,k;
int ans;
int xi[N],yi[N],c[N];//1白0黑 
int numt[2*K],numf[2*K];//每个y坐标上满足条件的数量 (正确的，错误的) 
void insert(){
	in(n),in(k);
	for(int i=1;i<=n;i++){
		int x,y;
		char ci;
		in(x),in(y);
		cin>>ci;
		xi[i]=x%(2*k),yi[i]=y%(2*k);
		if(ci=='W')c[i]=1;
		else c[i]=0;
	}
} 
int ex_max(int x,int y,int z){
	return max(x,max(y,z));
} 
void work(){
	for(int i=0;i<=k;i++){
		int tot=0;
		memset(numt,0,sizeof(numt));
		memset(numf,0,sizeof(numf));
		for(int j=1;j<=n;j++){
			int pdx=0,pdy=0;//覆盖即为1 (两条k粗的线划分网格 
			if(xi[j]>=i&&xi[j]<i+k)pdx=1;
			if(yi[j]>=0&&yi[j]<k)pdy=1;
			if(pdx^pdy==c[j])tot++,numt[yi[j]]++;
			else numf[yi[j]]++;
		}
		for(int j=0;j<k;j++){
//			cout<<tot<<endl;
//			system("pause");
			ans=ex_max(ans,tot,n-tot);
			tot=tot-numt[j]+numf[j]-numt[j+k]+numf[j+k];
		}
		ans=ex_max(ans,tot,n-tot);
	}
}
int main(){
	insert();
	work();
	cout<<ans;
}