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
#include<bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define ll long long
ll gcd(int x, int y) { return (x % y)? gcd(y, x % y): y; }    //最大公約数
ll lcm(ll x, ll y) { return x / gcd(x, y) * y; }     //最小公倍数
using Graph = vector<vector<int>>;
ll inf=300000000000000000;
const double PI = 3.14159265358979323846;
int main(){
  int n,k;
  cin >> n >> k;
  vector<vector<int>> x(1+k,vector<int>(1+k,0));
  vector<vector<int>> y(k+1,vector<int>(k+1,0));
  rep(i,n){
    int a,b;
    char c;
    cin >> a >> b >> c;
    int o=(a/k)+(b/k);
    if((o%2==0&&c=='B')||(o%2==1&&c=='W'))x[1+a%k][1+b%k]++;
    else y[a%k+1][b%k+1]++;
  }
  rep(i,k){
    rep(j,k){
      x[i+1][j+1]=x[i+1][j+1]+x[i+1][j]+x[i][j+1]-x[i][j];
      y[i+1][j+1]=y[i+1][j+1]+y[i+1][j]+y[i][j+1]-y[i][j];
      
    }
  }
  int ans=0;
  rep(i,k+1)rep(j,k+1)ans=max(ans,x[i][j]*2+x[k][k]-x[i][k]-x[k][j]+y[i][k]+y[k][j]-2*y[i][j]);
  rep(i,k+1)rep(j,k+1){
    ans=max(ans,y[i][j]*2+y[k][k]-y[i][k]-y[k][j]+x[i][k]+x[k][j]-2*x[i][j]);
    //if(ans==5)cout << i << " " << j << " " << y[i][j] << " " << y[k][k] << " " << y[i][k] << " " << y[k][
  }
  cout << ans << endl;
}
