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 <iostream>
#include <vector>
using namespace std;

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);

  int N, K; cin >> N >> K;
  int K2 = K * 2;
  
  vector<vector<int>> a(K2, vector<int>(K2));
  for (int i = 0; i < N; i++) {
    int x, y; cin >> x >> y;
    char c; cin >> c;
    if (c == 'W') x += K;
    a[x%K2][y%K2]++;
  }

  vector<vector<int>> cs(K2+1, vector<int>(K2+1));
  for (int i = 0; i < K2; i++) {
    for (int j = 0; j < K2; j++) {
      cs[i+1][j+1] = cs[i][j+1] + cs[i+1][j]
                   - cs[i][j] + a[i][j];
    }
  }

  auto query = [&](int x1, int y1, int x2, int y2) {
    return cs[x2][y2] - cs[x1][y2] - cs[x2][y1] + cs[x1][y1];
  };

  int ans = 0;
  for (int i = 0; i < K; i++) {
    for (int j = 0; j < K; j++) {
      int res = query(0, 0, i, j)
              + query(i, j, i+K, j+K)
              + query(0, j+K, i, K2)
              + query(i+K, 0, K2, j)
              + query(i+K, j+K, K2, K2);
      ans = max(ans, res);
      ans = max(ans, N-res);
    }
  }
  cout << ans << "\n";
}