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 <bits/stdc++.h>
#define _CRT_SECURE_NO_WARNINGS
#define ll long long
#define BUF 50
#define INF 1<<30
using namespace std;
ll MOD = 1e9 + 7;
ll A, B, C, D, G, H, N, M, L, K, P, Q, R, W, X, Y, Z;
string S;
ll ans = 0;
template<typename T>
struct RectangleSum {
	vector<vector<T>>imos;

	RectangleSum(vector<vector<T>>in) :imos(in.size(), vector<T>(in.size(), 0)) {
		for (int i = 1; i < in.size(); i++) {
			for (int j = 1; j < in.size(); j++) {
				imos[i][j] = in[i][j] + imos[i - 1][j] + imos[i][j - 1] - imos[i - 1][j - 1];
			}
		}
	}

	////xy平面を見て(p1,p2)が右上
	T GetRectangle(T p1, T p2, T q1, T q2) {
		return imos[p1][p2] - imos[q1 - 1][p2] - imos[p2][q2 - 1] + imos[q1 - 1][q2 - 1];
	}
};

int main() {
	cin >> N >> K;
	vector<vector<int>>recb(K + 1, vector<int>(K + 1, 0));
	vector<vector<int>>recw(K + 1, vector<int>(K + 1, 0));
	for (int i = 0; i < N; i++) {
		int x, y;
		char c;
		cin >> x >> y >> c;
		if ((x / K + y / K) & 1) {
			if (c == 'B')recw[x%K + 1][y%K + 1]++;
			else recb[x % K + 1][y % K + 1]++;
		}
		else {
			if (c == 'B')recb[x%K + 1][y%K + 1]++;
			else recw[x%K + 1][y%K + 1]++;
		}
	}
	RectangleSum<int> b(recb);
	RectangleSum<int> w(recw);
	ans = max(b.GetRectangle(K, K, 1, 1), w.GetRectangle(K, K, 1, 1));
	for (int i = 1; i < K; i++) {
		ll tmp1 = b.GetRectangle(i, K, 1, 1) + w.GetRectangle(K, K, i + 1, 1);
		ll tmp2 = w.GetRectangle(i, K, 1, 1) + b.GetRectangle(K, K, i + 1, 1);
		ll tmp3 = b.GetRectangle(K, i, 1, 1) + w.GetRectangle(K, K, 1, i + 1);
		ll tmp4 = w.GetRectangle(K, i, 1, 1) + b.GetRectangle(K, K, 1, i + 1);
		ans = max(ans, max(max(tmp1, tmp2), max(tmp3, tmp4)));
	}
	for (int i = 1; i < K; i++) {
		for (int j = 1; j < K; j++) {
			ll tmp = b.GetRectangle(i, j, 1, 1) + b.GetRectangle(K, K, i + 1, j + 1) + w.GetRectangle(K, j, i + 1, 1) + w.GetRectangle(i, K, 1, j + 1);
			ans = max(tmp, ans);
		}
	}
	cout << ans << endl;
}