Problem: 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:

cba927b2484fad94fb5ff7473e9aadef.png

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.

Examples

Input

4 3
0 1 W
1 2 W
5 3 B
5 4 B


Output

4


Input

2 1000
0 0 B
0 1 W


Output

2


Input

6 2
1 2 B
2 1 W
2 2 B
1 0 B
0 6 W
4 5 W


Output

4

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pb push_back
#define fi first
#define se second
typedef pair<ll,ll> P;
#define Mod 1000000007
using VP = vector<P>; using VVP = vector<VP>;
using VI = vector<ll>; using VVI = vector<VI>; using VVVI = vector<VVI>;
#define INF 1ll<<59

VVI w(2000,VI(2000,0));
ll s;

ll ww(ll a,ll b){
  if(a>=0&&b>=0) return w[a][b];
  else return 0;
}

ll solve(ll a,ll b,ll c, ll d){
  return ww(c,d)+ww(a,b)-ww(c,b)-ww(a,d);
}

int main(){
  ll i,j;
  ll n,k;
  cin>>n>>k;
  
  ll x,y;
  s=2*k;
  VVI v(s,VI(s,0));
  char c;
  for(i=0;i<n;i++){
    cin>>x>>y>>c;
    if(c=='B') x+=k;
    v[x%s][y%s]++;
  }
  for(j=0;j<s;j++) w[0][j]=v[0][j];

  for(i=1;i<s;i++) {
    for(j=0;j<s;j++){
      w[i][j]=w[i-1][j]+v[i][j];
    }
  }
  for(j=1;j<s;j++){
    for(i=0;i<s;i++){
      w[i][j]+=w[i][j-1];
    }
  }
  ll ans=0;
  for(i=0;i<k;i++){
    for(j=0;j<k;j++){
      ll temp=0;
      temp+=solve(-1,-1,i,j)+solve(i,j,i+k,j+k)+solve(i+k,-1,s-1,j);
      temp+=solve(-1,j+k,i,s-1)+solve(i+k,j+k,s-1,s-1);
      ans=max(ans,temp);
      ans=max(ans,n-temp);
    }
  }
  cout<<ans<<endl;

  return 0;
}