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<stdlib.h>
 
int main(){
 
  int N, K, i, j, k, m, a, b, num, max;
  char color;
  int x[100000], y[100000];
  char c[100000];
  scanf("%d %d", &N, &K);
  for(i=0; i<N; i++){
    scanf(" %d %d %c", &a, &b, &color);
    x[i]=a; y[i]=b; c[i]=color;
  }
 
  num=0; max=0;
  for(i=0; i<N; i++){
  
    for(j=1-K; j<=0; j++){
      for(k=1-K; k<=0; k++){
        num=1;
        for(m=0;m<N;m++){
          if(m==i)continue;
          a=x[i]+j; b=y[i]+k;
          if(x[m]< a && y[m]< b){a+=(K-1);b+=(K-1);}
          if(x[m]< a && y[m]>=b){a+=(K-1);         }
          if(x[m]>=a && y[m]< b){         b+=(K-1);}
          if(((abs(a-x[m])/K)%2)^((abs(b-y[m])/K)%2)){
            if(c[m]!=c[i]){num++;/*printf("m=%d\n", m);*/}
          }else{
            if(c[m]==c[i]){num++;/*printf("m=%d\n", m);*/}
          }
        }
        /*printf("i=%d, a=%d, b=%d, num=%d\n", i, a, b, num);*/
        if(num>max)max=num;
      }
    }
 
  }
 
  printf("%d\n", max);
 
}