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: Runtime Error)

#include<stdio.h>
#include<stdlib.h>

struct data{
  int x,y;
  char c;
};

int N,K;

int xcomp(const void *p1,const void *p2){
  struct data n1,n2;
  n1=*(const struct data *)p1;
  n2=*(const struct data *)p2;
  if(n1.x<n2.x) return -1;
  else if(n1.x>n2.x) return 1;
  else{
    if(n1.y<n2.y) return -1;
    else return 1;
  }
  return 0;
}

int ycomp(const void *p1,const void *p2){
  struct data n1,n2;
  n1=*(const struct data *)p1;
  n2=*(const struct data *)p2;
  if(n1.y<n2.y) return -1;
  else if(n1.y>n2.y) return 1;
  else{
    if(n1.x<n2.x) return -1;
    else return 1;
  }
  return 0;
}

int check(struct data p,int x,int y){
  if(p.x<x) p.x+=2*K;
  if(p.y<y) p.y+=2*K;
  if(p.c=='W'){
    if(x<=p.x && p.x<x+K && y<=p.y && p.y<y+K) return 1;
    else if(x+K<=p.x && p.x<x+2*K && y+K<=p.y && p.y<y+2*K) return 1;
  }else{
    if(x<=p.x && p.x<x+K && y+K<=p.y && p.y<y+2*K) return 1;
    else if(x+K<=p.x && p.x<x+2*K && y<=p.y && p.y<y+K) return 1;
  }
  return 0;
}

int main(void){
  int **DP,KK;
  int i,j,k,x,y,ind[2],indy[2],max;
  char c;
  struct data *p;
  
  scanf("%d%d%*c",&N,&K);
  KK=2*K;
  DP=(int **)calloc(K+1,sizeof(int *));
  for(i=0;i<=K;i++) DP[i]=(int *)calloc(K+1,sizeof(int));
  p=(struct data*)calloc(N,sizeof(struct data));

  // 代入
  for(i=0;i<N;i++){
    scanf("%d%d%*c%c%*c",&x,&y,&c);
    x%=KK;
    y%=KK;
    p[i]=(struct data){x,y,c};
  }

  //Sort by x
  qsort(p,N,sizeof(struct data),xcomp);

  //puts("x-sorted");

  //DP00;
  DP[0][0]=0;
  ind[0]=ind[1]=0;
  for(i=0;i<N;i++){
    DP[0][0]+=check(p[i],0,0);
    if(ind[1]==0 && i!=0 && p[i].x>=K && p[i-1].x<K) ind[1]=i;
  }

  //x-DP
  for(i=0;i<K;i++){
    DP[i+1][0]=DP[i][0];
    while(p[ind[0]].x==i){
      if(check(p[ind[0]],i,0)==1) DP[i+1][0]--;
      else DP[i+1][0]++;
      ind[0]++;
    }
    while(p[ind[1]].x==i+K){
      if(check(p[ind[1]],i,0)==0) DP[i+1][0]++;
      else DP[i+1][0]--;
      ind[1]++;
    }
  }

  //Sort by y
  qsort(p,N,sizeof(struct data),ycomp);
  ind[0]=ind[1]=0;
  for(i=0;i<N;i++){
    if(ind[1]==0 && i!=0 && p[i].y>=K && p[i-1].y<K) ind[1]=i;
  }
  indy[0]=ind[0];
  indy[1]=ind[1];

  //puts("y-sorted");

   //y-DP
  for(i=0;i<=K;i++){
    ind[0]=indy[0];
    ind[1]=indy[1];
    for(j=0;j<K;j++){
      DP[i][j+1]=DP[i][j];
      while(p[ind[0]].y==j){
	if(check(p[ind[0]],i,j)==1) DP[i][j+1]--;
	else DP[i][j+1]++;
	ind[0]++;
      }
      while(p[ind[1]].y==j+K){
	if(check(p[ind[1]],i,j)==0) DP[i][j+1]++;
	else DP[i][j+1]--;
	ind[1]++;
      }
    }
  }

  max=DP[0][0];
  for(i=0;i<=K;i++){
    for(j=0;j<=K;j++){
      if(DP[i][j]>max) max=DP[i][j];
      //printf("%d%c",DP[i][j],j==K?'\n':' ');
    }
  }
  printf("%d\n",max);

  for(i=0;i<=K;i++) free(DP[i]);
  free(DP);
  free(p);
  return 0;
  
}