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)

#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
#define _GLIBCXX_DEBUG
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
using namespace std;
using ll = long long;
const int INF = 1001001001;
const int MOD = 1000000007;
typedef long long ll;

template<typename T> void print(const T & v);


int main() {
    ll N;
    cin >> N;
    vector<ll> t(N), x(N), y(N);

    for (int i = 0; i < N; i++) {
        cin >> t.at(i) >> x.at(i) >> y.at(i);
    }
    ll a = 0, b = 0, ten = 0;
    bool ans = true;
    for (int i = 0; i < N; i++) {
        ll c = abs(x.at(i)-a) + abs(y.at(i)-b);
        ll d = t.at(i) - ten;
        if(d < c){
            cout << "No" << endl;
            return 0;
        }
        if(d == c){
            a = x.at(i);
            b = y.at(i);
            ten = t.at(i);
        }else if(d >= c && (d-c) % 2 == 0){
            a = x.at(i);
            b = y.at(i);
            ten = t.at(i);
        } else {
            cout << "No" << endl;
            return 0;
        }
    }
    cout << "Yes" << endl;
    return 0;
}


// Use For Debug
template <typename T>
void print(T const & v){
    for(int i = 0; i < v.size(); i++){
        if(i) cout << " ";
        cout << v[i];
    }
    cout << endl;
}