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>
using namespace std;
using ll=long long;
using datas=pair<ll,ll>;
using ddatas=pair<double,double>;
using tdata=pair<ll,datas>;
using vec=vector<ll>;
using mat=vector<vec>;
using pvec=vector<datas>;
using pmat=vector<pvec>;
#define For(i,a,b) for(i=a;i<(ll)b;i++)
#define bFor(i,a,b) for(i=a;i>=(ll)b;i--)
#define rep(i,N) For(i,0,N)
#define rep1(i,N) For(i,1,N)
#define brep(i,N) bFor(i,N-1,0)
#define all(v) (v).begin(),(v).end()
#define allr(v) (v).rbegin(),(v).rend()
#define vsort(v) sort(all(v))
#define vrsort(v) sort(allr(v))
#define endl "\n"
#define pb push_back
#define output(v) do{bool f=0;for(auto outi:v){cout<<(f?" ":"")<<outi;f=1;}cout<<endl;}while(0)
const ll mod=1000000007;
const ll inf=1LL<<60;
template<class T> inline bool chmax(T& a,T b){bool x=a<b;if(x)a=b;return x;} 
template<class T> inline bool chmin(T& a,T b){bool x=a>b;if(x)a=b;return x;} 

double distance(ddatas& x,ddatas& y){
  double a=x.first-y.first,b=x.second-y.second;
  return sqrt(a*a+b*b);
}

ll modinv(ll a) {
  ll b=mod,u=1,v=0,t;
  while(b){
    t=a/b;
    a-=t*b; swap(a,b);
    u-=t*v; swap(u,v);
  }
  return (u+mod)%mod;
}

ll moddevide(ll a,ll b){return (a*modinv(b))%mod;}

vec modncrlistp,modncrlistm;

ll modncr(ll n,ll r){
  ll i,size=modncrlistp.size();
  if(size<=n){
    modncrlistp.resize(n+1);
    modncrlistm.resize(n+1);
    if(!size){
      modncrlistp[0]=modncrlistm[0]=1;
      size++;
    }
    For(i,size,n+1){
      modncrlistp[i]=modncrlistp[i-1]*i%mod;
      modncrlistm[i]=modinv(modncrlistp[i]);
    }
  }
  return modncrlistp[n]*modncrlistm[r]%mod*modncrlistm[n-r]%mod;
}

ll modpow(ll a,ll n){
  ll res=1;
  while(n){
    if(n&1)res=res*a%mod;
    a=a*a%mod;
    n>>=1;
  }
  return res;
}

ll gcd(ll a,ll b){if(!b)return a;return (a%b==0)?b:gcd(b,a%b);}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}

ll countdigits(ll n){
  ll ans=0;
  while(n){n/=10;ans++;}
  return ans;
}

ll sumdigits(ll n){
  ll ans=0;
  while(n){ans+=n%10;n/=10;}
  return ans;
}

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  ll i,j,N,K,ans=0,res=1;
  cin>>N>>K;
  mat w(K+2,vec(K+2,0)),b(K+2,vec(K+2,0));
  string s;
  while(N--){
    cin>>i>>j>>s;
    res=i/K+j/K+(s[0]=='B');
    if(res%2) b[i%K+1][j%K+1]++;
    else w[i%K+1][j%K+1]++;
  }
  K++;
  rep(i,K){
    rep(j,K){
      b[i][j+1]+=b[i][j];
      w[i][j+1]+=w[i][j];
    }
  }
  rep(i,K){
    rep(j,K){
      b[i+1][j]+=b[i][j];
      w[i+1][j]+=w[i][j];
    }
  }
  rep(i,K+1)rep(j,K+1){
    res=b[K][K]-b[i][K]-b[K][j]+b[i][j]*2+w[i][K]+w[K][j]-w[i][j];
    chmax(ans,res);
    res=w[K][K]-w[i][K]-w[K][j]+w[i][j]*2+b[i][K]+b[K][j]-b[i][j];
    chmax(ans,res);
  }
  cout<<ans<<endl;
  return 0;
}