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 <bits/stdc++.h>
 
#define M_PI       3.14159265358979323846
 
using namespace std;
 
//conversion
//------------------------------------------
inline int toInt(string s) { int v; istringstream sin(s); sin >> v; return v; }
template<class T> inline string toString(T x) { ostringstream sout; sout << x; return sout.str(); }
inline int readInt() { int x; scanf("%d", &x); return x; }
 
//typedef
//------------------------------------------
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef pair<int, PII> TIII;
typedef long long LL;
typedef unsigned long long ULL;
typedef vector<LL> VLL;
typedef vector<VLL> VVLL;
 
 
//container util
 
//------------------------------------------
#define ALL(a)  (a).begin(),(a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define MP make_pair
#define SZ(a) int((a).size())
#define SQ(a) ((a)*(a))
#define EACH(i,c) for(typeof((c).begin()) i=(c).begin(); i!=(c).end(); ++i)
#define EXIST(s,e) ((s).find(e)!=(s).end())
#define SORT(c) sort((c).begin(),(c).end())
 
//repetition
//------------------------------------------
#define FOR(i,s,n) for(int i=s;i<(int)n;++i)
#define REP(i,n) FOR(i,0,n)
#define MOD 1000000007
 
#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define trav(a, x) for(auto& a : x)
#define all(x) x.begin(), x.end()
#define sz(x) (int)(x).size()
 
 
 
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
const double EPS = 1E-8;
 
#define chmin(x,y) x=min(x,y)
#define chmax(x,y) x=max(x,y)
const int INF = 100000000;
 
struct Edge {
    int to, from;
    ll cost;
    Edge(int from, int to, ll cost): from(from), to(to), cost(cost) {}
};
 
class UnionFind {
public:
    vector <ll> par; 
    vector <ll> siz; 
 
    UnionFind(ll sz_): par(sz_), siz(sz_, 1LL) {
        for (ll i = 0; i < sz_; ++i) par[i] = i;
    }
    void init(ll sz_) {
        par.resize(sz_);
        siz.assign(sz_, 1LL);
        for (ll i = 0; i < sz_; ++i) par[i] = i;
    }
 
    ll root(ll x) { 
        while (par[x] != x) {
            x = par[x] = par[par[x]];
        }
        return x;
    }
 
    bool merge(ll x, ll y) {
        x = root(x);
        y = root(y);
        if (x == y) return false;
        if (siz[x] < siz[y]) swap(x, y);
        siz[x] += siz[y];
        par[y] = x;
        return true;
    }
 
    bool issame(ll x, ll y) { 
        return root(x) == root(y);
    }
 
    ll size(ll x) { 
        return siz[root(x)];
    }
};
 
typedef vector<vector<Edge>> AdjList;
AdjList graph;
 
ll mod_pow(ll x, ll n, ll mod){
    ll res = 1;
    bool c = false;
    while(n){
        if(n&1) res = res * x;
 
        if(res > mod){
            c = true;
            res %= mod;
        }
        x = x * x %mod;
        n >>= 1;
    }
    if(c) return mod;
    return res;
}
 
#define SIEVE_SIZE 5000000+10
bool sieve[SIEVE_SIZE];
void make_sieve(){
    for(int i=0; i<SIEVE_SIZE; ++i) sieve[i] = true;
    sieve[0] = sieve[1] = false;
    for(int i=2; i*i<SIEVE_SIZE; ++i) if(sieve[i]) for(int j=2; i*j<SIEVE_SIZE; ++j) sieve[i*j] = false;
}
 
bool isprime(ll n){
    if(n == 0 || n == 1) return false;
    for(ll i=2; i*i<=n; ++i) if(n%i==0) return false;
    return true;
}
 
template<typename T>
vector<T> gauss_jordan(const vector<vector<T>>& A, const vector<T>& b){
    int n = A.size();
    vector<vector<T>> B(n, vector<T>(n+1));
 
    for(int i=0; i<n; ++i){
        for(int j=0; j<n; ++j){
            B[i][j] = A[i][j];
        }
    }
 
    for(int i=0; i<n; ++i) B[i][n] = b[i];
 
    for(int i=0; i<n; ++i){
        int pivot = i;
        for(int j=i; j<n; ++j){
            if(abs(B[j][i]) > abs(B[pivot][i])) pivot = j;
        }
        swap(B[i], B[pivot]);
 
        if(abs(B[i][i]) < EPS) return vector<T>(); //解なし
 
        for(int j=i+1; j<=n; ++j) B[i][j] /= B[i][i];
        for(int j=0; j<n; ++j){
            if(i != j){
                for(int k=i+1; k<=n; ++k) B[j][k] -= B[i][j] * B[i][k];
            }
        }
    }
 
    vector<T> x(n);
 
    for(int i=0; i<n; ++i) x[i] = B[i][n];
    return x;
    
}
 
 
 
 
typedef vector<ll> vec;
typedef vector<vec> mat;
 
mat mul(mat &A, mat &B) {
    mat C(A.size(), vec((int)B[0].size()));
    for(int i=0; i<A.size(); ++i){
        for(int k=0; k<B.size(); ++k){
            for(int j=0; j<B[0].size(); ++j){
                C[i][j] = (C[i][j] + A[i][k] * B[k][j] %MOD) % MOD;
            }
        }
    }
    return C;
}
mat mat_pow(mat A, ll n) {
    mat B(A.size(), vec((int)A.size()));
 
    for(int i=0; i<A.size(); ++i){
        B[i][i] = 1;
    }
 
    while(n > 0) {
        if(n & 1) B = mul(B, A);
        A = mul(A, A);
        n >>= 1;
    }
    return B;
}
 
bool operator<(const pii& a, const pii& b){
    if(a.first == b.first) return a.second < b.second;
    return a.first < b.first;
}
 
const int MAX = 510000;
long long fac[MAX], finv[MAX], inv[MAX];
 
// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
 
// 二項係数計算
long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
 
int bit[1000010];
int sums(int i){
    i++;
    int s = 0;
    while(i > 0){
        s += bit[i];
        i -= i & -i;
    }
    return s;
}
void add(int i, int x){
    i++;
    while(i <= 1000010){
        bit[i] += x;
        i += i & -i;
    }
}
long long extGCD(long long a, long long b, long long &x, long long &y) {
    if (b == 0) {
        x = 1;
        y = 0;
        return a;
    }
    long long d = extGCD(b, a%b, y, x);
    y -= a/b * x;
    return d;
}
 
ll GCD(ll a, ll b){
    
    if(b == 0) return a;
    return GCD(b, a%b);
}

int sum[2010][2010];
int grid[2010][2010];

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(18);
    
    int N, K; cin >> N >> K;
    REP(i, N){
        int x, y; char c; cin >> x >> y >> c;

        if(c == 'W'){
            y += K;
        }

        x %= (2*K);
        y %= (2*K);
        grid[x][y] = 1;
    }

    for(int i=0; i<2*K; i++){
        for(int j=0; j<2*K; j++){
            sum[i+1][j+1] = sum[i][j+1] + sum[i+1][j] - sum[i][j] + grid[i][j];
        }
    }

    
    int ans = 0;

    for(int i=0; i<=K; i++){
        for(int j=0; j<=K; j++){
            int tmp = 0;
            tmp += sum[i][j];
            tmp += sum[i][2*K] - sum[i][j+K];
            tmp += sum[i+K][j+K] - sum[i+K][j] - sum[i][j+K]   + sum[i][j];
            tmp += sum[2*K][j] - sum[i+K][j];
            tmp += sum[2*K][2*K] - sum[2*K][j+K] - sum[i+K][2*K] + sum[i+K][j+K];

            ans = max(ans, max(tmp, N-tmp));
        }
    }
    cout << ans << endl;

    return 0;
}