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<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>
#include<algorithm>
#include<cstring>
#include<string>
#include<cassert>
#include<cmath>
#include<climits>
#include<iomanip>
#include<bitset>
#include<unordered_map>

using namespace std;

#define REP(i,n) for(ll (i)=0;(i)<(n);(i)++)
#define rep(i,j,n) for(ll (i)=(j);(i)<(n);(i)++)
#define FOR(i,c) for(decltype((c).begin())i=(c).begin();i!=(c).end();++i)
#define ll long long
#define ull unsigned long long
#define all(hoge) (hoge).begin(),(hoge).end()
typedef pair<ll, ll> P;
const long long INF = 1LL << 60;
const long long MOD = 1e9 + 7;
typedef vector<ll> Array;
typedef vector<Array> Matrix;


template<class T> inline bool chmin(T& a, T b) {
	if (a > b) {
		a = b;
		return true;
	}
	return false;
}
template<class T> inline bool chmax(T& a, T b) {
	if (a < b) {
		a = b;
		return true;
	}
	return false;
}

//グラフ関連
struct Edge {//グラフ
	ll to, cap, rev;
	Edge(ll _to, ll _cap, ll _rev) {
		to = _to; cap = _cap; rev = _rev;
	}
};
typedef vector<Edge> Edges;
typedef vector<Edges> Graph;

void add_edge(Graph& G, ll from, ll to, ll cap, bool revFlag, ll revCap) {
	G[from].push_back(Edge(to, cap, (ll)G[to].size()));
	if (revFlag)G[to].push_back(Edge(from, revCap, (ll)G[from].size() - 1));
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	cout.tie(nullptr);

	ll n, k;
	cin >> n >> k;

	Matrix mp(2 * k, Array(2 * k, 0));
	REP(i, n) {
		ll x, y;
		char c;
		cin >> x >> y >> c;
		if (c == 'W') y += k;
		mp[x % (2 * k)][y % (2 * k)]++;
	}

	

	Matrix sum(2 * k + 1, Array(2 * k + 1,0));
	REP(i, 2 * k) {
		REP(j, 2 * k) {
			sum[i + 1][j + 1] = mp[i][j]+sum[i + 1][j] + sum[i][j + 1] - sum[i][j];
		}
	}
	auto ssum = [&](ll x1, ll y1,ll x2,ll y2)->ll {
		ll ans = sum[x2 + 1][y2 + 1] - sum[x2 + 1][y1] - sum[x1][y2 + 1] + sum[x1][y1];
		if (ans > 0) return ans;
		else return 0;
	};

	ll ans=0;
	REP(i,k+1) {
		REP(j,k+1) {
			ll con = 0;
			con += ssum(0, 0, i - 1, j - 1);
			con += ssum(i, j, i + k - 1, j + k - 1);
			con += ssum(i + k, j + k, 2 * k - 1, 2 * k - 1);
			con += ssum(0, j + k, i - 1, 2 * k-1);
			con += ssum(i + k, 0, j - 1, 2 * k - 1);
			chmax(ans, con);
		}
	}
	cout << ans;


	return 0;
}