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 <vector>
#include<algorithm>
#include<iomanip>
using namespace std;
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define REP(i,m,n) for(int i=m;i<(int)(n);i++)
typedef pair<int,int> pint;
typedef long long ll;
const int inf   = 1<<30;
const ll  longinf = 1LL<<60;
int dx[4]={1,0,-1,0}, dy[4]={0,1,0,-1};
 
int main(){
	int n,k,l;
	scanf("%d %d",&n,&k);
	l=2*k;
	int x[n],y[n],table[k+1][k+1];
	bool black[n];
	char c[n];
	rep(i,k+1)rep(j,k+1)table[i][j]=0;
	rep(i,n){
		scanf("%d %d %c",&x[i],&y[i],&c[i]);
		x[i]%=l;
		y[i]%=l;
		black[i]=(c[i]=='B');
		if(x[i]>=k){
			x[i]-=k;
			black[i]=!black[i];
		}
		if(y[i]>=k){
					y[i]-=k;
					black[i]=!black[i];
		}
		if(black[i]){
			table[0][0]++;
			table[0][y[i]+1]--;
			table[x[i]+1][0]--;
			table[x[i]+1][y[i]+1]+=2;
		}
		else{
			table[0][y[i]+1]++;
			table[x[i]+1][0]++;
			table[x[i]+1][y[i]+1]-=2;
		}
	}
	rep(i,k)REP(j,1,k)table[i][j]+=table[i][j-1];
	rep(i,k)REP(j,1,k)table[j][i]+=table[j-1][i];
	int ans=0;
	rep(i,k)rep(j,k){
		table[i][j]=max(table[i][j],n-table[i][j]);
		ans=max(ans,table[i][j]);
	}
	cout<<ans<<endl;
  return 0;
  }