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> //C++の標準ライブラリを一行で一括でインクルードする
#include <math.h> //数学関数と数学定数を利用する
#define rep(i,n) for (int i = 0; i < (n); i++)
using namespace std;
typedef long long ll;
template<class T> void chmax(T &a,T b) { if (a<b) a=b;}
template<class T> void chmin(T &a,T b) { if (a>b) a=b;}
 
ll memo[2][1005][1005] = {0};
char c;
ll x,y,bw,n,k;
 
ll solve(){
  ll ans = 1, now[2];
  rep(i,k)rep(j,k)rep(l,2){
    now[l] = memo[l][i][j];
    if(i != k-1 && j != k-1){
      now[l] += memo[l][k-1][k-1];
      now[l] -= memo[l][i][k-1];
      now[l] -= memo[l][k-1][j];
      now[l] += memo[l][i][j];
    }
    now[l] += memo[1-l][k-1][j] - memo[1-l][i][j];
    now[l] += memo[1-l][i][k-1] - memo[1-l][i][j];
    ans = max(ans, now[l]);
  }
  return ans;
}
 
int main(){
  cin >> n >> k;
  for(int i = 0; i < n; i++){
    cin >> x >> y >> c;
    bw = (c == 'B');
    memo[(x/k + y/k + bw)%2][x%k][y%k]++;
  }
  
  rep(l,2)rep(i,k)rep(j,k){
    if(j != 0) memo[l][i][j] += memo[l][i][j-1];
    if(i != 0) memo[l][i][j] += memo[l][i-1][j];
    if(i != 0 && j != 0){
      memo[l][i][j] -= memo[l][i-1][j-1];
    }
  }
  
  cout << solve() << endl;
 
  return 0;
}