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;
typedef long long lint;
typedef pair<int,int> P;
const int    INF=(int)1e9;
const int    MOD=(int)1e9+7;
const double EPS=(double)1e-10;
struct Accelerate_Cin{
    Accelerate_Cin(){
        cin.tie(0); ios::sync_with_stdio(0);cout<<fixed<<setprecision(20);
    };
};

int  N,K;
int  board[2010][2010];

signed main(){
  cin>>N>>K;

  int sum=0,white=0;
  for(int t=1;t<=N;t++){
    int x,y;char c;cin>>x>>y>>c;
    x%=2*K,y%=2*K;
    if(c=='B') board[x+1][y+1]++,sum++;
    if(c=='W') board[x+1][y+1]--,sum--,white++;
  }

  for(int i=1;i<=2*K;i++){
    for(int j=1;j<=2*K;j++){
      board[i][j]+=board[i][j-1];
    }
  }

  for(int j=1;j<=2*K;j++){
    for(int i=1;i<=2*K;i++){
      board[i][j]+=board[i-1][j];
    }
  }

  int mx=-INF;
  for(int i=0;i<=K;i++){
    for(int j=0;j<=K;j++){
      int a,b,c,d,e;
      a=board[i][j];
      b=board[i+K][j+K]-board[i+K][j]-board[i][j+K]+board[i][j];
      c=board[i][2*K]-board[i][j+K];
      d=board[2*K][j]-board[i+K][j];
      e=board[2*K][2*K]-board[2*K][j+K]-board[i+K][2*K]+board[i+K][j+K];
      int temp=a+b+c+d+e;
      mx=max(mx,max(temp,sum-temp));
    }
  }
  mx+=white;
  cout<<mx<<"\n";
  return 0;
}