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;

const int N = 4005;

int i, j, n, k, x, y, a[N][N], rs;
char c;

int getSquare(int x, int y, int xx, int yy) {
  if (xx < 0 || yy < 0) return 0;
  int ans = a[xx][yy];
  if (x > 0) ans -= a[x - 1][yy];
  if (y > 0) ans -= a[xx][y - 1];
  if (x > 0 && y > 0) ans += a[x - 1][y - 1];
  return ans;
}

int main() {
  ios_base::sync_with_stdio(0);

  cin >> n >> k;
  for (i = 1; i <= n; ++i) {
    cin >> x >> y >> c;
    if (c == 'B') y += k;
    x %= 2 * k;
    y %= 2 * k;
    ++a[x + 1][y + 1];
  }

  for (i = 1; i < N; ++i)
    for (j = 1; j < N; ++j)
      a[i][j] += a[i - 1][j] + a[i][j - 1] - a[i - 1][j - 1];

  for (i = 1; i <= 2 * k; ++i)
    for (j = 1; j <= 2 * k; ++j) {
      int aux = getSquare(i - k + 1, j - k + 1, i, j);
      aux += getSquare(i - 2 * k + 1, j - 2 * k + 1, i - k, j - k);
      aux += getSquare(i - 2 * k + 1, j + 1, i - k, j + k);
      aux += getSquare(i + 1, j - 2 * k + 1, i + k, j - k);
      aux += getSquare(i + 1, j + 1, i + k, j + k);
      aux += getSquare(i - k + 1, j + k + 1, i, j + 2 * k);
      aux += getSquare(i + k + 1, j - k + 1, i + 2 * k, j);
      rs = max(rs, aux);
    }

  cout << rs << '\n';
  
  return 0;
}