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<cstdio>
#include<algorithm>
#define MAXN 100000
#define MAXK 1000
using namespace std;
struct node
{
	int x,y,opt;
}poi[MAXN+5];
int dp[MAXK+5][MAXK+5],ctr[MAXK+5][MAXK+5];
int n,k;
int main()
{
	scanf("%d %d",&n,&k);
	for(int i=1;i<=n;i++)
	{
		int a,b,tot=0,optnum;
		char opt;
		scanf("%d %d %c",&a,&b,&opt);
		if(opt=='B')		optnum=1;
		else if(opt=='W')	optnum=0;
		tot+=a/k; a%=k;
		tot+=b/k; b%=k;
		poi[i].x=a; poi[i].y=b;
		if(tot%2==1)		poi[i].opt=(optnum+1)%2;
		else				poi[i].opt=optnum;
	}
	for(int i=1;i<=n;i++)
	{
		int x=poi[i].x,y=poi[i].y,opt=poi[i].opt;
		if(poi[i].opt==1)
			ctr[k][k]++,ctr[x][k]--,ctr[k][y]--,ctr[x][y]+=2;
		else
			ctr[x][k]++,ctr[k][y]++,ctr[x][y]-=2;
	}
	int ans=0;
	for(int y=k;y>=0;y--)
		for(int x=k;x>=0;x--)
		{
			dp[x][y]=dp[x][y+1]+dp[x+1][y]-dp[x+1][y+1]+ctr[x][y];
			ans=max(ans,max(dp[x][y],n-dp[x][y]));
		}
	printf("%d\n",ans);
	return 0;
}
/*
4 3
5 3 B
0 1 W
1 2 W
5 4 B

*/