Problem:
Score : 500 points

Problem Statement

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:

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.

Sample Input 1

4 3
0 1 W
1 2 W
5 3 B
5 4 B

Sample Output 1

4

He can satisfy all his desires by painting as shown in the example above.

Sample Input 2

2 1000
0 0 B
0 1 W

Sample Output 2

2

Sample Input 3

6 2
1 2 B
2 1 W
2 2 B
1 0 B
0 6 W
4 5 W

Sample Output 3

4

Solution in C (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <string.h>
 
int main(){
  int n,k,i;
  scanf("%d %d",&n,&k);
  int x[100005],y[100005];
  char c[100005];
  for(i=0;i<n;i++){
    scanf("%d %d %s",&x[i],&y[i],&c[i]);
    x[i] = x[i]%(2*k);
    if(c[i]=='W'){
      y[i] = (y[i]+k)%(2*k);
    }else{
      y[i] = y[i]%(2*k);
    }
  }
  int ans=0,last=0,xx,yy;
  int a,b,eee,aaa;
  for(a=0;a<2*k;a++){
    for(b=0;b<2*k;b++){
      for(i=0;i<n;i++){
        xx = x[i]-a;
        if(xx<0){
          xx += 2*k;
        }
        yy = y[i]-b;
        if(yy<0){
          yy += 2*k;
        }
        if((xx <= k-1 && yy <= k-1) || (xx >= k && yy >= k)){
          ans += 1;
        }
        if(ans>last){
          last = ans;
          eee = a;
          aaa = b;
        }
      }
      ans = 0;
    }
  }
  printf("%d\n",last);
  return 0;
}