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 implements Runnable {

    private static Scanner sc = new Scanner(System.in);

    public static void main(String[] args) {
        new Thread(null, new Main(), "", 128 * 1024 * 1024).start();
    }

    static int[][] cumSum;

    public void run() {
        try (Scanner in = new Scanner(System.in)) {
            int n = in.nextInt(), k = in.nextInt(), k2 = k * 2;
            int[][] w = new int[k2][k2];
            for (int i = 0; i < n; i++) {
                int x = in.nextInt(), y = in.nextInt();
                if (in.next().charAt(0) == 'B') {
                    y += k;
                }
                w[y % k2][x % k2]++;
            }

            cumSum = new int[k2 + 1][k2 + 1];
            for (int i = 1; i <= k2; i++) {
                for (int j = 1; j <= k2; j++) {
                    cumSum[i][j] = cumSum[i - 1][j] + cumSum[i][j - 1] - cumSum[i - 1][j - 1] + w[i - 1][j - 1];
                }
            }

            int max = 0;
            for (int i = 0; i < k2; i++) {
                for (int j = 0; j < k2; j++) {
                    int y0 = k2, x0 = k2;
                    int y1 = Math.min(i + k, k2), x1 = Math.min(j + k, k2);
                    int y2 = i, x2 = j;
                    int y3 = Math.max(i - k, 0), x3 = Math.max(j - k, 0);
                    int y4 = 0, x4 = 0;

                    int sum = sum(y1, y0, x1, x0)
                              + sum(y1, y0, x3, x2)
                              + sum(y2, y1, x2, x1)
                              + sum(y2, y1, x4, x3)
                              + sum(y3, y2, x3, x2)
                              + sum(y3, y2, x1, x0)
                              + sum(y4, y3, x4, x3)
                              + sum(y4, y3, x2, x1);
                    max = Math.max(sum, max);
                }
            }
            System.out.println(max);
        }
    }

    private int sum(int ya, int yb, int xa, int xb) {
        return cumSum[ya][xa] + cumSum[yb][xb] - cumSum[yb][xa] - cumSum[ya][xb];
    }
}