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: Accepted)

#include <algorithm>
#include <array>
#include <climits> // INT_MAX, LONG_MAX, LLONG_MAX, UINT_MAX, ULONG_MAX, ULLONG_MAX
#include <cmath>
#include <cstdint>
#include <cstring> // memset
#include <iostream>
#include <map>
#include <set>
#include <string>
#include <vector>
 
#define ALL(x) (x).begin(), (x).end()
#define INF 1e9
#define LOG(x) cout << #x << " = " << (x) << endl
#define REP(i, n) for (int i = 0; i < (int)(n); ++i)
 
using namespace std;

int s[2][1000][1000];


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

    REP(i, n) {
        int xi, yi;
        string ci;
        cin >> xi >> yi >> ci;

        if (ci[0] == 'B')
            xi += k; // Do not use -= for %
        if (xi % (2 * k) >= k) {
            xi += k; // Do not use -= for %
            yi += k; // Do not use -= for %
        }

        if (yi % (2 * k) >= k)
            s[1][xi % k][yi % k]++;
        else
            s[0][xi % k][yi % k]++;
    }

    REP(p, 2) REP(q, k) REP(r, k)
        s[p][q][r] += (q > 0 ? s[p][q - 1][r] : 0) + (r > 0 ? s[p][q][r - 1] : 0) - (q > 0 && r > 0 ? s[p][q - 1][r - 1] : 0);

    int best = 0;

    REP(p, k) REP(q, k) {
        best = max(best,
                   max(s[0][k - 1][k - 1] - s[0][p][k - 1] - s[0][k - 1][q] + 2 * s[0][p][q] + s[1][p][k - 1] + s[1][k - 1][q] - 2 * s[1][p][q],
                       s[1][k - 1][k - 1] - s[1][p][k - 1] - s[1][k - 1][q] + 2 * s[1][p][q] + s[0][p][k - 1] + s[0][k - 1][q] - 2 * s[0][p][q]));
    }
    
    cout << best << endl;
    
    return 0;
}