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: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
int map[2000][1000];
void printm(int n,  int k){
  int i, j;
  printf("\n%d:\n",n);
  for(i=0;i<2*k;i++){
    for(j=0;j<2*k;j++) printf("%d ",map[j][2*k-1-i]);
    printf("\n");
  }
}

void mapping(int k, int x, int y){
  int j;
  for(j=0;j+y<k;j++){
    map[x][(y+j)%(2*k)]++; map[(x+k)%(2*k)][(y+j)%(2*k)]--;
    if(x>=k) map[0][(y+j)%(2*k)]++;
  }
  for(j=k-y;j<k;j++){
    map[(x+k)%(2*k)][(y+j-k)%(2*k)]++; map[(x)%(2*k)][(y+j-k)%(2*k)]--;
    if(x<k) map[0][(y+j-k)%(2*k)]++;
  }
}

int main(void){
  int n, k, i, j, x, y, r=0, max=0;
  char c;
  scanf("%d %d",&n,&k);
  for(i=0;i<2*k;i++) for(j=0;j<k;j++) map[i][j]=0; 
  for(i=0;i<n;i++){
    scanf("%d %d %c",&x,&y,&c);
    r=0;
    if(((int)x/k)%2==1) r++;
    if(((int)y/k)%2==1) r++;
    x%=k;  y%=k;
    if((r%2==1 && c=='W') || (r%2==0 && c=='B')){
      mapping(k, x+k, y);
    }else{
      mapping(k, x, y);
    }
    
    //printm(i,k);
  }
  for(i=0;i<k;i++){
    for(j=0;j<2*k-1;j++){
      map[j+1][i]+=map[j][i];
      if(map[j+1][i]>max) max=map[j+1][i];
    }
  }
  
					     
  printf("%d\n",max);
  return 0;
}