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<math.h>
typedef struct{
	int x;
	int y;
	char c;
}block;

int main(void){
	int n, k;
	char cc;
	block block1[100001];
	scanf("%d%d", &n, &k);
	for(int i = 0; i < n; i++){
		scanf("%d%d%c%c", &block1[i].x, &block1[i].y, &cc, &block1[i].c);
	}
	int max = 0, div_x = 0, div_y = 0, max_ans = 0;
	for(int j = 0; j <= k; j++){
		for(int z = 0; z <= 2 * k; z++){
		
			for(int i = 0; i < n; i++){
				div_x =abs( (block1[i].x + j )/ k);
				div_y = abs((block1[i].y + z )/ k);
				int mod = (div_x + div_y)%2;
				if( mod == 0 && block1[i].c == 'W'){
//					printf("%d %d (%d, %d)\n", j, z, block1[i].x, block1[i].y);
					max++;
				}
				if(mod == 1 &&block1[i].c == 'B'){
//					printf("%d %d (%d, %d)\n", j, z, block1[i].x, block1[i].y);
					max++;
				}
			}
			if(max_ans < max)max_ans = max;
			max = 0;
		}
	}
	printf("%d\n", max_ans);
}