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 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int main()
{
	int i, N, K, x, y, count[2][2000][2000] = {};
	char c;
	scanf("%d %d", &N, &K);
	for (i = 1; i <= N; i++) {
		scanf("%d %d %c", &x, &y, &c);
		if (c == 'B') count[0][x%(K*2)][y%(K*2)]++;
		else count[1][x%(K*2)][y%(K*2)]++;
	}
	
	int j, ans = 0, tmp, diff[2000] = {};
	for (x = 0; x < K; x++) {
		for (y = 0; y < K; y++) {
			diff[x] -= count[0][x][y] - count[1][x][y];
			ans += count[0][x][y];
		}
		for (; y < K * 2; y++) {
			diff[x] += count[0][x][y] - count[1][x][y];
			ans += count[1][x][y];
		}
	}
	for (; x < K * 2; x++) {
		for (y = 0; y < K; y++) {
			diff[x] += count[0][x][y] - count[1][x][y];
			ans += count[1][x][y];
		}
		for (; y < K * 2; y++) {
			diff[x] -= count[0][x][y] - count[1][x][y];
			ans += count[0][x][y];
		}
	}
	for (x = 0, tmp = ans; x < K; x++) {
		tmp += diff[x] + diff[x+K];
		if (tmp > ans) ans = tmp;
	}
	for (; x < K * 2; x++) {
		tmp -= diff[x] + diff[x-K];
		if (tmp > ans) ans = tmp;
	}
	for (y = 0; y < K; y++) {
		for (x = 0; x < K; x++) {
			diff[x] += count[0][x][y] - count[1][x][y] - count[0][x][y+K] + count[1][x][y+K];
			tmp -= count[0][x][y] - count[1][x][y] - count[0][x][y+K] + count[1][x][y+K];
		}
		for (; x < K * 2; x++) {
			diff[x] -= count[0][x][y] - count[1][x][y] - count[0][x][y+K] + count[1][x][y+K];
			tmp += count[0][x][y] - count[1][x][y] - count[0][x][y+K] + count[1][x][y+K];
		}
		for (x = 0; x < K; x++) {
			tmp += diff[x] + diff[x+K];
			if (tmp > ans) ans = tmp;
		}
		for (; x < K * 2; x++) {
			tmp -= diff[x] + diff[x-K];
			if (tmp > ans) ans = tmp;
		}
	}
	for (; y < K * 2; y++) {
		for (x = 0; x < K; x++) {
			diff[x] += count[0][x][y] - count[1][x][y] - count[0][x][y-K] + count[1][x][y-K];
			tmp -= count[0][x][y] - count[1][x][y] - count[0][x][y-K] + count[1][x][y-K];
		}
		for (; x < K * 2; x++) {
			diff[x] += count[0][x][y] - count[1][x][y] - count[0][x][y-K] + count[1][x][y-K];
			tmp -= count[0][x][y] - count[1][x][y] - count[0][x][y-K] + count[1][x][y-K];
		}
		for (x = 0; x < K; x++) {
			tmp += diff[x] + diff[x+K];
			if (tmp > ans) ans = tmp;
		}
		for (; x < K * 2; x++) {
			tmp -= diff[x] + diff[x-K];
			if (tmp > ans) ans = tmp;
		}
	}
	
	printf("%d\n", ans);
	fflush(stdout);
	return 0;
}