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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn=4000;
int n,k,wt[maxn+10][maxn+10],blk[maxn+10][maxn+10],ans;
char s[5];
int main(){
	scanf("%d%d",&n,&k);
	for(int i=1;i<=n;++i){
		int x,y; scanf("%d%d",&x,&y); scanf("%s",s);
		x%=k*2; y%=k*2; ++x; ++y; if(s[0]=='W') ++wt[x][y]; else ++blk[x][y];
	}
	for(int i=k*2+1;i<=k*4;++i)
		for(int j=k*2+1;j<=k*4;++j) wt[i][j]=wt[i-k*2][j-k*2],blk[i][j]=blk[i-k*2][j-k*2];
	for(int i=1;i<=k*4;++i)
		for(int j=1;j<=k*4;++j){
			wt[i][j]+=wt[i-1][j]+wt[i][j-1]-wt[i-1][j-1];
			blk[i][j]+=blk[i-1][j]+blk[i][j-1]-blk[i-1][j-1];
		}
	for(int i=0;i<k;++i)
		for(int j=0;j<k;++j){
			int t=wt[i+k][j+k]-wt[i][j+k]-wt[i+k][j]+wt[i][j];
			t+=wt[i+k*2][j+k*2]-wt[i+k][j+k*2]-wt[i+k*2][j+k]+wt[i+k][j+k];
			t+=blk[i+k][j+k]-blk[i][j+k]-blk[i+k][j]+blk[i][j];
			t+=blk[i+k*2][j+k*2]-blk[i+k][j+k*2]-blk[i+k*2][j+k]+blk[i+k][j+k];
			ans=max(ans,max(t,n-t));
		}
	printf("%d",ans); return 0;
}