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 <istream>
#include <cstdio>
#include <vector>
#include <cmath>
#include <algorithm>
#include <map>
#include <set>
//#include <unordered_map>
#include <cstring>
#include <string.h>
#include <iomanip>
#include <queue>
#include <stack>
#include <list>
#include <sstream>
//#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> ii;
typedef vector<int> vi;
typedef long double ld;

#define ALL(V) V.begin(), V.end()
#define SZ(V) (int)V.size()
#define pb push_back
#define mp make_pair
#define f first
#define s second

#define fopen freopen("substrcmp.in", "r", stdin); freopen("substrcmp.out","w", stdout)
#define FOR(i, a, b) for(int i = (a); i < (b); ++i)
#define FORR(i, b, a) for(int i = (b) - 1; i >= (a); --i)
#define FILL(A,value) memset(A,value,sizeof(A))
#define IOSFAST ios::sync_with_stdio(false); cin.tie(0)

const ll MOD = 1000LL * 1000 * 1000 + 7;
const int INF = 1e9;
const ll base = 1e18;
const int MAX = 1000;
const double EPS = 1e-9;
const double PI = acos(-1.);

int n,k;
int t[4000][4000];

void mark(int x1, int y1)
{
	int x2 = x1+k;
	int y2 = y1+k;
	//cout << x1 << ' ' << y1 << ' ' << x2 <<' ' << y2 << endl;
	FOR(i,y1,y2)
	{
		t[i+2000][x1+2000]++;
		t[i+2000][x2+2000]--;
	}
}

int main()
{
	cin >> n >> k;
	FOR(i,0,n)
	{
		int x,y;
		char ch;
		cin >> x >> y >> ch;
		int oldx = x/k;
		int oldy = y/k;
		x%=2*k;
		y%=2*k;
		if (ch == 'B')
		{
			mark(0-x,0-(y-k));
			mark(0-y,0-(x-k));
		} else
		{
			x%=k;
			y%=k;
			mark(-x,-y);
			x+=k;
			y+=k;
			mark(-x,-y);
		}
	}
	int cur = 0;
	int ans = -INF;
	FOR(i,0,4000)
	{
		FOR(j,0,4000)
		{
			cur+=t[i][j];
			ans = max(ans,cur);
		}
	}
	cout << ans;
	return 0;
}