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
#include<cstdio>
#include<cstring>
#include<algorithm>
#define MAXN 2000
using namespace std;

int n,k,ans;
int a[MAXN+5][MAXN+5];

void Add(int x,int y,int ex,int ey)
{
	++a[x][y];
	++a[ex][ey];
	--a[x][ey];
	--a[ex][y];
}

int main()
{
	scanf("%d%d",&n,&k);
	for(int i=1;i<=n;i++)
	{
		int x,y;char c;
		scanf("%d %d %c",&x,&y,&c);
		x %= 2*k;y %= 2*k;
		if(((x<k&&y<k)||(x>=k&&y>=k))==(c=='B'))
		{
			int nx=x % k,ny=y % k;
			Add(0,0,nx+1,ny+1);
			Add(nx+k+1,0,2*k,ny+1);
			Add(0,ny+k+1,nx+1,2*k);
			Add(nx+k+1,ny+k+1,2*k,2*k);
			Add(nx+1,ny+1,nx+k+1,ny+k+1);
		}
		else
		{
			int nx=x % k,ny=y % k;
			Add(nx+1,0,nx+k+1,ny+1);
			Add(nx+1,ny+k+1,nx+k+1,2*k);
			Add(0,ny+1,nx+1,ny+k+1);
			Add(nx+k+1,ny+1,2*k,ny+k+1);
		}
	}
	for(int i=1;i<MAXN;i++)
        a[0][i]+=a[0][i-1];
	for(int i=1;i<MAXN;i++)
    {
        a[i][0]+=a[i-1][0];
		for(int j=1;j<MAXN;j++)
            a[i][j]+=a[i-1][j]-a[i-1][j-1]+a[i][j-1];
    }
	for(int i=0;i<2*k;i++)
		for(int j=0;j<2*k;j++)
			ans=max(ans,a[i][j]);
	printf("%d",ans);
}
