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;

int main() {
  int n, k;
  cin >> n >> k;
  vector<int> x(n), y(n);
  vector<char> c(n);

  for (int i = 0; i < n; i++) {
    cin >> x[i] >> y[i] >> c[i];
  }
  for (int i = 0; i < n; i++) {
    if (c[i] == 'W') {
      x[i] += k;
    }
    x[i] %= 2 * k;
    y[i] %= 2 * k;
  }

  vector<vector<int> > vec(2*k + 5, vector<int>(2*k + 5, 0));
  for (int i = 0; i < n; i++) {
    vec[x[i]+1][y[i]+1]++;
  }

  for (int i = 1; i < 2*k + 1; i++) {
    for (int j = 1; j < 2*k + 1; j++) {
      vec[i][j] += vec[i-1][j];
    }
  }
  for (int i = 1; i < 2*k + 1; i++) {
    for (int j = 1; j < 2*k + 1; j++) {
      vec[i][j] += vec[i][j-1];
    }
  }

  int ans = -1e9;
  for (int i = 0; i <= k; i++) {
    for (int j = 0; j <= k; j++) {
      int leftup = vec[i][j];
      int center = vec[i+k][j+k] - vec[i][j+k] - vec[i+k][j] + vec[i][j];
      int leftdw = vec[2*k][j] - vec[i+k][j];
      int rightup = vec[i][2*k] - vec[i][j+k];
      int rightdw = vec[2*k][2*k] - vec[2*k][j+k] - vec[i+k][2*k] + vec[i+k][j+k];
      int sum = leftup + center + leftdw + rightup + rightdw;
      ans = max(ans, max(sum, n-sum));
    }
  }
  cout << ans << endl;
  return 0;
}