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 (Clang 3.8.0) (judge verdict: Accepted)

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <functional>

using namespace std;

int s(vector<vector<int>>& f, int bx, int by, int ex, int ey) {
	if (bx < 0 || by < 0 || ex >= f.size() || ey >= f[0].size()) return 0;

	return f[bx][by] - f[bx][ey] - f[ex][by] + f[ex][ey];
}

int main() {
	int n, k;
	cin >> n >> k;

	vector<vector<int>> f(2*k+1, vector<int>(2*k+1, 0));
	int x, y;
	char c;

	for (int i = 0; i< n; i++) {
		cin >> x >> y >> c;
		if (c == 'W') {
			x += k;
		}
		x = x % (2*k);
		y = y % (2*k);
		f[x+1][y+1] += 1;
	}

	for (int i = 1; i < 2*k+1; i++) {
		for (int j = 1; j < 2*k+1; j++) {
			f[i][j] += f[i][j-1] + f[i-1][j] - f[i-1][j-1];
		}
	}

	int ans = 0;
	for (int i = 0; i < k; i++) {
		for (int j = 0; j < k; j++) {
			int c1 = s(f, 0, 0, i+1, j+1);
			int c2 = s(f, i+k+1, 0, 2*k, j+1);
			int c3 = s(f, i+1, j+1, i+k+1, j+k+1);
			int c4 = s(f, 0, j+k+1, i+1, 2*k);
			int c5 = s(f, i+k+1, j+k+1, 2*k, 2*k);
			int cb = c1 + c2 + c3 + c4 + c5;
			if (cb > ans) {
				ans = cb;
			}
			int c6 = s(f, i+1, 0, i+k+1, j+1);
			int c7 = s(f, i+1, j+k+1, i+k+1, 2*k);
			int c8 = s(f, 0, j+1, i+1, j+k+1);
			int c9 = s(f, i+k+1, j+1, 2*k, j+k+1);
			int cb2 = c6 + c7 + c8 + c9;
			if (cb2 > ans) {
				ans = cb2;
			}
		}
	}
	cout << ans << endl;
	return 0;
}