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;

const int MAXK = 1000;

int c[2*MAXK+1][2*MAXK+1];
int sum[2*MAXK+1][2*MAXK+1];

int query(int x1, int y1, int x2, int y2){
  if(x1 > x2 || y1 > y2) return 0;
  // return sum of x1 <= x <= x2 && y1 <= y <= y2
  return sum[x2][y2] - sum[x1 - 1][y2] - sum[x2][y1 - 1] + sum[x1 - 1][y1 - 1];
}

int main(){
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  int N, K; cin >> N >> K;
  for(int i=0;i<N;i++){
    int x, y;
    char color;
    cin >> x >> y >> color;
    if(color == 'B'){
      c[x %(2*K) + 1][y %(2*K) + 1]++;
    }else{
      c[x %(2*K) + 1][(y + K) %(2*K) + 1]++;
    }
  }

  for(int i=1;i<=2*K;i++){
    for(int j=1;j<=2*K;j++){
      sum[i][j] = c[i][j] + sum[i-1][j] + sum[i][j-1] - sum[i-1][j-1];
      // cout << sum[i][j] << " ";
    }
    // cout << endl;
  }

  int maxnum = 0;
  for(int x=1;x<=K+1;x++){
    for(int y=1;y<=K+1;y++){
      // count num in five black squares
      int num = query(1, 1, x-1, y-1) + query(1, y+K, x-1, 2*K)
      + query(x, y, x+K-1, y+K-1) + query(x+K, 1, 2*K, y-1) + query(x+K, y+K, 2*K, 2*K);
      maxnum = max(max(num, N-num), maxnum);
    }
  }
  cout << maxnum << endl;
}