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 <bits/stdc++.h>
using namespace std;
#define gc getchar()
inline int read(){
	int x=0;char c=gc;
	for(;c<48||c>57;c=gc)
		if(c=='W')return 0;
		else if(c=='B')return 1;
	for(;c>47&&c<58;c=gc)
		x=x*10+c-48;
	return x;
}
#define io read()

#define maxn 4005
int m,k,n,n2;
int w[maxn][maxn],b[maxn][maxn];

inline void init(){
	m=io;k=io;
	n=k<<1;n2=n<<1;
	while(m--){
		int x=io%n+1,y=io%n+1;
		++(io?b:w)[x][y];
	}
	for(int i=1;i<=n;++i)
		for(int j=1;j<=n;++j){
			b[n+i][j]=b[i][n+j]=b[n+i][n+j]=b[i][j];
			w[n+i][j]=w[i][n+j]=w[n+i][n+j]=w[i][j];
		}
	for(int i=1;i<=n2;++i)
		for(int j=1;j<=n2;++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];
		}
}

inline int sum(int a[][maxn],int x,int y){
	--x;--y;
	return a[x+k][y+k]-a[x][y+k]-a[x+k][y]+a[x][y];
}
inline void solve(){
	int ans=0;
	for(int i=1;i<=n;++i)
		for(int j=1;j<=n;++j){
			int tmp=sum(w,i,j)+sum(b,i+k,j)+sum(b,i,j+k)+sum(w,i+k,j+k);
			if(tmp>ans)ans=tmp;
		}
	printf("%d\n",ans);
}

int main(){
	init();
	solve();
	return 0;
}