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: Time Limit Exceeded)

#include <bits/stdc++.h>

using namespace std;

#define _ << " " <<
#define pb push_back
#define ff first
#define ss second


typedef long long ll;
typedef long double ld;

//#define int ll

struct Point {
    int x, y;
    char col;
};

int n, k;

int get(int x, int y, char c, vector<Point> &a) {
    int ans = 0;
    for (int i = 0; i < n; ++i) {
        int tmpx = (a[i].x - abs(abs(a[i].x - x) - k) % k);
        int tmpy = (a[i].y - abs(abs(a[i].y - y) - k) % k);
        /*if (x == 0 && y == 1 && c == 'W') {
            cout << (abs(x - tmpx) % k + abs(y - tmpy) % k) _ tmpx _ tmpy << endl;
        }*/
        if (((abs(x - tmpx) + abs(y - tmpy)) / k) % 2 == 0) {
            if (a[i].col == c) {
                ans++;
            }
        } else {
            if (a[i].col != c) {
                ++ans;
            }
        }
    }
    //if (x == 0 && y == 1 && c == 'W') cout << ans << endl;
    return ans;
}

void solve() {
    cin >> n >> k;
    vector<Point> a(n);
    for (int i = 0; i < n; ++i) {
        cin >> a[i].x >> a[i].y >> a[i].col;
    }
    int ans = 0;
    for (int i = 0; i < k; ++i) {
        for (int j = 0; j < k; ++j) {
            ans = max(ans, max(get(i, j, 'B', a), get(i, j, 'W', a)));
        }
    }
    cout << ans;
}

signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    //freopen("dominoes.in", "r", stdin);
    //freopen("dominoes.out", "w", stdout);
    int test = 1;
    //cin >> test;
    while (test--) {
        solve();
    }
}