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 <math.h>
#include <string.h>
#include <stdlib.h>

#define llong long long
#define fr(i,l,r) for(i=(l);i<(r);i++)
#define min(p,q) ((p)<(q)?(p):(q))
#define max(p,q) ((p)>(q)?(p):(q))

int main(void)
{
	//変数の宣言
	int n,m;
	int x,y;
	char c;
	static int map[2010][2010];
	static int ruiseki[2010][2010];
	static int ruisekitemp[2010][2010];
	
	//よく使う変数
	int i,j,k,l;
	int flag=0;
	int ans=0;
	int count=0;
	int sum=0;
	int temp,temp1,temp2;
	//int max,min;
	int len;
	//データの読み込み
	
	scanf("%d %d",&n,&m);
	
	for(i=0;i<=2000;i++){
		for(j=0;j<=2000;j++){
			map[i][j]=0;
		}
	}
	
	for(i=0;i<n;i++){
		scanf("%d %d %c",&x,&y,&c);
		if(c=='W') x=x+m;
		x=x%(2*m);
		y=y%(2*m);
		map[x][y]++;
	}
	
//	printf("nは%dです\n", n);
//	printf("データの読み込み終了\n");
	//実際の処理
	
	//二次元累積和の計算
	
	for(i=0;i<2*m;i++){
		for(j=0;j<2*m;j++){
			if(j==0) ruisekitemp[i][0]=map[i][0];
			ruisekitemp[i][j]=ruisekitemp[i][j-1]+map[i][j];
		}
	}
	
	for(j=0;j<2*m;j++){
		for(i=0;i<2*m;i++){
			if(i==0) ruiseki[0][j]=ruisekitemp[0][j];
			ruiseki[i][j]=ruiseki[i-1][j]+ruisekitemp[i][j];
		}
	}
	
	for(i=0;i<m;i++){
		for(j=0;j<m;j++){
			temp=0;
			temp=temp+ruiseki[i+m][j+m]-ruiseki[i+m][j]-ruiseki[i][j+m]+ruiseki[i][j];
			temp=temp+ruiseki[i][j];
			temp=temp+ruiseki[2*m-1][2*m-1]-ruiseki[i+m][2*m-1]-ruiseki[2*m-1][j+m]+ruiseki[i+m][j+m];
			temp=temp+ruiseki[2*m-1][j]-ruiseki[i+m][j];
			temp=temp+ruiseki[i][2*m-1]-ruiseki[i][j+m];
			if(max(temp,n-temp)>ans) ans=max(temp,n-temp);
		}
	}
	
	#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>

#define llong long long
#define fr(i,l,r) for(i=(l);i<(r);i++)
#define min(p,q) ((p)<(q)?(p):(q))
#define max(p,q) ((p)>(q)?(p):(q))

int main(void)
{
	//変数の宣言
	int n,m;
	int x,y;
	char c;
	static int map[2010][2010];
	static int ruiseki[2010][2010];
	static int ruisekitemp[2010][2010];
	
	//よく使う変数
	int i,j,k,l;
	int flag=0;
	int ans=0;
	int count=0;
	int sum=0;
	int temp,temp1,temp2;
	//int max,min;
	int len;
	//データの読み込み
	
	scanf("%d %d",&n,&m);
	
	for(i=0;i<=2000;i++){
		for(j=0;j<=2000;j++){
			map[i][j]=0;
		}
	}
	
	for(i=0;i<n;i++){
		scanf("%d %d %c",&x,&y,&c);
		if(c=='W') x=x+m;
		x=x%(2*m);
		y=y%(2*m);
		map[x][y]=map[x][y]+1;
	}
	
//	printf("nは%dです\n", n);
//	printf("データの読み込み終了\n");
	//実際の処理
	
	//二次元累積和の計算
	
	for(i=0;i<2*m;i++){
		for(j=0;j<2*m;j++){
			if(j==0){
			ruisekitemp[i][0]=map[i][0];
			}else{
				ruisekitemp[i][j]=ruisekitemp[i][j-1]+map[i][j];
			}
		}
	}
	
	for(j=0;j<2*m;j++){
		for(i=0;i<2*m;i++){
			if(i==0){
				ruiseki[0][j]=ruisekitemp[0][j];
			}else{
				ruiseki[i][j]=ruiseki[i-1][j]+ruisekitemp[i][j];
			}
		}
	}
	
	for(i=0;i<m;i++){
		for(j=0;j<m;j++){
			temp=0;
			temp=temp+ruiseki[i+m][j+m]-ruiseki[i+m][j]-ruiseki[i][j+m]+ruiseki[i][j];
			temp=temp+ruiseki[i][j];
			temp=temp+ruiseki[2*m-1][2*m-1]-ruiseki[i+m][2*m-1]-ruiseki[2*m-1][j+m]+ruiseki[i+m][j+m];
			temp=temp+ruiseki[2*m-1][j]-ruiseki[i+m][j];
			temp=temp+ruiseki[i][2*m-1]-ruiseki[i][j+m];
			if(temp<n-temp) temp=n-temp;
			if(temp>ans) ans=temp;
		}
	}
	
	
	
	
//	printf("計算部分終了\n");
	//出力
	
	printf("%d",ans);
	
//	printf("結果の出力終了\n");
	
	return 0;
	
}


	
	
//	printf("計算部分終了\n");
	//出力
	
	printf("%d",ans);
	
//	printf("結果の出力終了\n");
	
	return 0;
	
}