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<cstdio>
using namespace std;
int a[2002][2002];
int b[2002][2002];
char s[4];
inline int f(int (*d)[2002],int sx,int sy,int ex,int ey){
	return d[ex][ey]-d[ex][sy]-d[sx][ey]+d[sx][sy];
}
int main()
{
	int i,j,n,m;
	scanf("%d%d",&n,&m);
	while(n--){
		scanf("%d%d%s",&i,&j,s);
		if(s[0]=='W')a[i%(m+m)+1][j%(m+m)+1]++;
		else b[i%(m+m)+1][j%(m+m)+1]++;
	}
	for(i=1;i<=m+m;i++)for(j=1;j<=m+m;j++){
		a[i][j]+=a[i-1][j]+a[i][j-1]-a[i-1][j-1];
		b[i][j]+=b[i-1][j]+b[i][j-1]-b[i-1][j-1];
	}
	int ans=0,cnt=0;
	for(i=1;i<=m;i++)for(j=1;j<=m;j++){
		cnt=f(a,0,0,i,j)
		+f(b,i,0,i+m,j)
		+f(a,i+m,0,m+m,j)
		+f(b,0,j,i,j+m)
		+f(a,i,j,i+m,j+m)
		+f(b,i+m,j,m+m,j+m)
		+f(a,0,j+m,i,m+m)
		+f(b,i,j+m,i+m,m+m)
		+f(a,i+m,j+m,m+m,m+m);
		ans=max(ans,cnt);
		cnt=f(b,0,0,i,j)
		+f(a,i,0,i+m,j)
		+f(b,i+m,0,m+m,j)
		+f(a,0,j,i,j+m)
		+f(b,i,j,i+m,j+m)
		+f(a,i+m,j,m+m,j+m)
		+f(b,0,j+m,i,m+m)
		+f(a,i,j+m,i+m,m+m)
		+f(b,i+m,j+m,m+m,m+m);
		ans=max(ans,cnt);
	}
	cout<<ans;
	return 0;
}