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 <bits/stdc++.h>
using namespace std;

int main(){
  int n,k;
  cin >> n >> k;
  int x[n],y[n];
  char c[n];
  for(int i=0;i<n;i++) cin >> x[i] >> y[i] >> c[i];
  for(int i=0;i<n;i++){
    if(c[i]=='W'){
      x[i]+=k;
    }
    x[i]%=2*k,y[i]%=2*k;
  }
  int a[2*k+1][2*k+1];
  for(int i=0;i<2*k+1;i++){
    for(int j=0;j<2*k+1;j++){
      a[i][j]=0;
    }
  }
  for(int i=0;i<n;i++){
    a[x[i]][y[i]]++;
  }
  int ans=0;
  int sum[4*k+1][4*k+1];
  for(int i=0;i<4*k+1;i++) sum[i][0]=0,sum[0][i]=0;
  for(int i=1;i<4*k+1;i++){
    for(int j=1;j<4*k+1;j++){
      sum[i][j]=sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1]+a[(i-1)%(2*k)][(j-1)%(2*k)];
    }
  }
  for(int i=0;i<2*k+1;i++){
    for(int j=0;j<2*k+1;j++){
      int point=sum[i+k][j+k]+sum[i][j]-sum[i][j+k]-sum[i+k][j];
      point+=(sum[i+2*k][j+2*k]+sum[i+k][j+k]-sum[i+2*k][j+k]-sum[i+k][j+2*k]);
      ans=max(ans,point);
    }
  }
  /*for(int i=0;i<4*k+1;i++){
    for(int j=0;j<4*k+1;j++){
      cout << sum[i][j] << " ";
    }
    cout << endl;
  }*/
  cout << ans << endl;
}