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: Accepted)

#include <stdio.h>

int B[4001][4001], W[4001][4001];

int main(void) {
	int N, K;
	int i, j, dx, dy;
	int answer = 0;
	if (scanf("%d%d", &N, &K) != 2) return 1;
	for (i = 0; i < N; i++) {
		int x, y, c;
		char sumipe[4];
		if (scanf("%d%d%3s", &x, &y, sumipe) != 3) return 1;
		c = sumipe[0];
		for (dy = 0; dy <= 1; dy++) {
			for (dx = 0; dx <= 1; dx++) {
				(c == 'B' ? B : W)[y % (K * 2) + (K * 2) * dy + 1][x % (K * 2) + + (K * 2) * dx + 1]++;
			}
		}
	}
	for (i = 1; i <= K * 4; i++) for (j = 1; j <= K * 4; j++) B[i][j] += B[i][j - 1];
	for (i = 1; i <= K * 4; i++) for (j = 1; j <= K * 4; j++) B[i][j] += B[i - 1][j];
	for (i = 1; i <= K * 4; i++) for (j = 1; j <= K * 4; j++) W[i][j] += W[i][j - 1];
	for (i = 1; i <= K * 4; i++) for (j = 1; j <= K * 4; j++) W[i][j] += W[i - 1][j];
	for (i = 0; i < K * 2; i++) {
		for (j = 0; j < K * 2; j++) {
			int w1 = W[i + K][j + K] - W[i][j + K] - W[i + K][j] + W[i][j];
			int w2 = W[i + K + K][j + K + K] - W[i + K][j + K + K] - W[i + K + K][j + K] + W[i + K][j + K];
			int b1 = B[i + K + K][j + K] - B[i + K][j + K] - B[i + K + K][j] + B[i + K][j];
			int b2 = B[i + K][j + K + K] - B[i][j + K + K] - B[i + K][j + K] + B[i][j + K];
			int score = w1 + w2 + b1 + b2;
			if (score > answer) answer = score;
		}
	}
	printf("%d\n", answer);
	return 0;
}