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)

// #include <bits/stdc++.h>
#include <iostream>
#include <vector>
#include <bitset>
#include <map>
#include <unordered_map>
#include <ctime>

using namespace std;

#define rep(i,n) REP(i,0,n)
#define REP(i,s,e) for(int i=(s); i<(int)(e); i++)
#define repr(i, n) REPR(i, n, 0)
#define REPR(i, s, e) for(int i=(int)(s-1); i>=(int)(e); i--)
#define pb push_back
#define all(r) r.begin(),r.end()
#define rall(r) r.rbegin(),r.rend()
#define fi first
#define se second

typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

const int INF = 1e9;
const ll MOD = 1e9 + 7;
double EPS = 1e-8;

const int MAX_N = 1e5 + 10;
int x[MAX_N], y[MAX_N], z[MAX_N];
const int MAX_K = 1001;
const int MAX_S = 3 * MAX_K + 5;
int s[MAX_S];
const int BASE = MAX_K + 1;
int prvX[MAX_N];
int sxs[MAX_N];

typedef bitset<MAX_N> BS;
unordered_map<BS, int> mp;

void f(int n, int k) {
    BS bs;
    rep(sx, k) {
        rep(i, n) bs[i] = (((x[i] + sx) / k) & 1);
        mp[bs] = sx;
    }
}

int main() {
    int n, k;
    cin >> n >> k;
    rep(i, n) {
        char c;
        cin >> x[i] >> y[i] >> c;
        if (c == 'W') z[i] = 1;
        else z[i] = 0;
    }
    f(n, k);
    srand((unsigned)time(NULL));
    while(mp.size() > 300) {
        int x = rand()%mp.size();
        auto it = mp.begin();
        rep(i, x) ++it;
        mp.erase(it);
    }
    int cnt = 0;    
    for (auto& p : mp) sxs[cnt++] = p.se;
    int ans = 0;
    rep(i, cnt) rep(c, 2) {
        int sx = sxs[i];
        // int idxmin = INF, idxmax = -INF;
        rep(i, n) {
            int X = (x[i] + sx) / k;
            int Y = y[i] / k;
            int ymin = Y * k - y[i];
            int ymax = ymin + k;
            if ((X + Y + c) % 2 == z[i]) {
                ymin += k;
                ymax += k;
            }
            s[ymin + BASE] += 1;
            s[ymax + BASE] -= 1;
            // idxmin = min(idxmin, ymin + BASE);
            // idxmax = max(idxmax, ymax + BASE);
        }
        // REP(i, idxmin, idxmax) s[i + 1] += s[i];
        // REP(i, idxmin, idxmax + 1) {
        //     ans = max(ans, s[i]);
        //     s[i] = 0;
        // }
        rep(i, MAX_S - 1) s[i+1] += s[i];
        rep(i, MAX_S) {
            ans = max(ans, s[i]);
            s[i] = 0;
        } 
    }
    cout << ans << endl;
    return 0;
}