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 n, k;


int arr[2001][2001];
int arr_doubled[4002][4002];
int get(int y, int x){ 
  return arr_doubled[y+k-1][x+k-1]+arr_doubled[y-1][x-1]-arr_doubled[y-1][x+k-1]-arr_doubled[y+k-1][x-1];
}
int main() {
  /*int n = 2;
  int m = 4;
  int sum = 0;
  for (int i=1; i<=n; i++){
    int x = m*(m+1)/2;
    sum += n*m*m-n*x+m*n-i*m*m+i*x-m*i+m*m-x+m;
  }
  cout << sum;*/
  cin >> n >> k;
 
  for (int i=0; i<n; i++){
    int x, y;
    char c;
    cin >> x >> y >> c;
    if (c == 'W'){
      x += k;
    }
   
    arr[y%(2*k)][x%(2*k)]++;
  }
  
  for (int i=1; i<=4*k; i++){
    for (int j=1; j<=4*k; j++){
      arr_doubled[i][j] = arr_doubled[i][j-1]+arr[(i-1)%(2*k)][(j-1)%(2*k)];
    }
  }

 
  for (int i=1; i<=4*k; i++){
    for (int j=2; j<=4*k; j++){
      arr_doubled[j][i] += arr_doubled[j-1][i];
    }
  }
  int biggest = -1;
  for (int i=1; i<=2*k; i++){
    for (int j=1; j<=2*k; j++){
      biggest = max(get(i, j)+get(i+k, j+k), biggest);
      
    }
  }
  cout << biggest;
  return 0;
}