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;
typedef long long ll;

const ll INF = 1e9;
const ll MOD = 1e9 + 7;

const ll MAX = 1e5;

vector<vector<int> > drw;

int cnt(int x1, int y1, int x2, int y2) {
  return drw[x2][y2] - drw[x1][y2] - drw[x2][y1] + drw[x1][y1];
}

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

  drw.resize(K2 + 1, vector<int>(K2 + 1, 0));
  for (int i = 0; i < K2; i++) {
    for (int j = 0; j < K2; j++) {
      drw[i + 1][j + 1] = drw[i][j + 1] + drw[i + 1][j] - drw[i][j] + list[i][j];
    }
  }

  int ans = -1e9;
  for (int i = 0; i <= K; i++) {
    for (int j = 0; j <= K; j++) {
      int res = cnt(0, 0, i, j)
       + cnt(i, j, i + K, j + K)
       + cnt(0, j + K, i, K2)
       + cnt(i + K, 0, K2, j)
       + cnt(i + K, j + K, K2, K2);
       ans = max(ans, res);
       ans = max(ans, N - res);
    }
  }
  cout << ans << endl;
  return 0;
}
