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 JAVA):

Solution:
import java.util.Scanner;

public class Main {
  public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);

    int N = scanner.nextInt();
    int K = scanner.nextInt();

    int count = 0;
    int[][] delta = new int[K][K];
    for (int i = 0; i < N; i++) {
      int x = scanner.nextInt();
      int y = scanner.nextInt();
      char c = scanner.next().charAt(0);
      int p = x / K + y / K;

      delta[x % K][y % K]++;
      if ((p % 2 == 0 && c == 'W') || (p % 2 == 1 && c == 'B')) {
        delta[x % K][y % K] -= 2;
        count++;
      }
    }

    for (int i = 0; i < K; i++) for (int j = 1; j < K; j++) delta[i][j] += delta[i][j - 1];
    for (int j = 0; j < K; j++) for (int i = 1; i < K; i++) delta[i][j] += delta[i - 1][j];

    int max = count;
    for (int i = 0; i <= K; i++) {
      for (int j = 0; j <= K; j++) {
        int c1 = count, c2 = count + delta[K - 1][K - 1];
        if (i > 0) {
          c1 += delta[i - 1][K - 1];
          c2 -= delta[i - 1][K - 1];
        }
        if (j > 0) {
          c1 += delta[K - 1][j - 1];
          c2 -= delta[K - 1][j - 1];
        }
        if (i > 0 && j > 0) {
          c1 -= 2 * delta[i - 1][j - 1];
          c2 += 2 * delta[i - 1][j - 1];
        }
        max = Math.max(max, c1);
        max = Math.max(max, c2);
      }
    }

    System.out.println(max);
  }
}