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 <iostream>
#include <algorithm>
#include <vector>
#include <string.h>
typedef long long ll;

#define MAX(x, y)	(((x) > (y))? (x) : (y))
#define MIN(x, y)	(((x) < (y))? (x) : (y))
#define ABS(x)		(((x) < 0)? (-1*(x)) : (x))
#define REPI(x,y)	for (int i=(x);i<(y);i++)
#define REPJ(x,y)	for (int j=(x);j<(y);j++)
#define MEMCLR(X)	memset((X),0,sizeof((X)))

using namespace std;
typedef vector<vector<pair<int, int>>> GRA;

int i,j,N,K,x,y;
string c;
int m;
int A[3000][3000];

int main() {
	cin >> N >> K;

	REPI(0,N) {
		cin >> x >> y >> c;
		if (c[0] == 'W') y += K;
		x %= (2*K);
		y %= (2*K);
		A[x][y]++;
		A[x+K][y+K]++;
		A[x+K][y]--;
		A[x][y+K]--;
	}
	REPI(1,2*K) REPJ(0,2*K) A[i][j] = A[i-1][j];
	REPI(0,2*K) REPJ(1,2*K) A[i][j] = A[i][j-1];
	REPI(0,2*K) REPJ(0,2*K) m = MAX(m,A[i][j]);

	cout << m << endl;

	return 0;
}