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: Wrong Answer)

#define _USE_MATH_DEFINES
#include <iostream>
#include <string>
#include <list>
#include <vector>
#include <queue>
#include <algorithm>
#include <climits>
#include <cstring>
#include <cmath>
#include <stack>
#include <iomanip>
#include <tuple>
#include <functional>
#include <cfloat>
#include <map>

#define int long long
#define CONTAINS(v,n) (find((v).begin(), (v).end(), (n)) != (v).end())
#define SORT(v) sort((v).begin(), (v).end())
#define RSORT(v) sort((v).rbegin(), (v).rend())
#define ARY_SORT(a, size) sort((a), (a)+(size))
#define MAX(a,b) (((a) > (b)) ? (a) : (b))
#define MIN(a,b) (((a) < (b)) ? (a) : (b))
#define REMOVE(v,a) (v.erase(remove((v).begin(), (v).end(), (a)), (v).end()))
#define REVERSE(v) (reverse((v).begin(), (v).end()))
#define LOWER_BOUND(v,a) (lower_bound((v).begin(), (v).end(), (a)))
#define UPPER_BOUND(v,a) (upper_bound((v).begin(), (v).end(), (a)))
#define REP(i, n) for (int (i) = 0; (i) < (n); (i)++)
#define CONTAINS_MAP(m, a) (m).find((a)) != m.end()
using namespace std;

int N, K;
int X[100010];
int Y[100010];
char C[100010];

int mas[1000][2000];
int syaku[2000];

signed main()
{
	cin >> N >> K;
	REP(i, N) cin >> X[i] >> Y[i] >> C[i];

	REP(i, N)
	{
		int x = X[i];
		int y = Y[i];
		x %= (K * 2);
		y %= (K * 2);
		if (x >= K)
		{
			x -= K;
			if (y >= K)
			{
				y -= K;
			}
			else
			{
				y += K;
			}
		}
		if (C[i] == 'W')
		{
			if (y >= K)
			{
				y -= K;
			}
			else
			{
				y += K;
			}
		}
		mas[x][y]++;
	}

	for (int y = 0; y < K * 2; y++)
	{
		int cnt = 0;
		for (int i = 0; i < K; i++)
		{
			int p = i;
			int q = y;
			cnt += mas[p][q];
		}
		syaku[y] = cnt;
	}

	int max = 0;
	for (int x = 0; x < K; x++)
	{
		int ans = 0;
		for (int i = 0; i < K; i++)
		{
			ans += syaku[i];
		}
		max = MAX(max, ans);

		for (int i = 1; i < K * 2; i++)
		{
			ans -= syaku[i - 1];
			int k = (i + K) % (K * 2);
			ans += syaku[k];
			max = MAX(ans, max);
		}

		for (int y = 0; y < K * 2; y++)
		{
			int cnt = 0;
			syaku[y] -= mas[x][y];

			int yy = (y + K) % (K * 2);

			syaku[y] += mas[x][yy];
		}
	}

	cout << max << endl;
}