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;
using ll=long long;
using P=pair<int,int>;

const int MAX=2011;

int n,k,xy[MAX][MAX],X[MAX][MAX];

void prepare(){
  for(int i=1;i<=2*k+2;i++)
    for(int j=1;j<=2*k+2;j++)
      X[i][j]=xy[i][j]+X[i-1][j]+X[i][j-1]-X[i-1][j-1];
}

int getsize(int x1,int y1,int x2,int y2){
  if(x1>x2 || y1>y2)
    return 0;
  else
    return X[x2][y2]-X[x1-1][y2]-X[x2][y1-1]+X[x1-1][y1-1];
}

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin>>n>>k;
  for(int i=0;i<n;i++){
    int x,y;
    char c;
    cin>>x>>y>>c;
    if(c=='W') y+=k;
    x%=(2*k);
    y%=(2*k);
    xy[x+1][y+1]++;
  }
  prepare();
  int ans=0;
  for(int i=1;i<=k;i++){
    for(int j=1;j<=k;j++){
      int tmp=getsize(1,1,i-1,j-1)+getsize(i,j,i+k-1,j+k-1)+getsize(i+k,j+k,2*k,2*k)+getsize(1,j+k,i-1,2*k)+getsize(i+k,1,2*k,j-1);
      ans=max(ans,max(tmp,n-tmp));
	}
  }
  cout<<ans<<endl;
  return 0;
}