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:

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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 sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();
        int[][] count = new int[4 * k + 1][4 * k + 1];

        for (int i = 0; i < n; i++) {
            int x = sc.nextInt();
            int y = sc.nextInt();
            char c = sc.next().charAt(0);
            // convert white into black to regard every wish to be black
            if (c == 'W') y += k;

            // want (x, y) to be black
            x %= 2 * k;
            y %= 2 * k;

            count[x][y]++;
            count[x][y + k]--;
            count[x + k][y]--;
            count[x + k][y + k] += 2;
            count[x + 2 * k][y + k]--;
            count[x + k][y + 2 * k]--;
            count[x + 2 * k][y + 2 * k]++;
        }

        for (int row = 0; row < count.length; row++) {
            for (int col = 1; col < count[0].length; col++) {
                count[col][row] += count[col - 1][row];
            }
        }
        for (int col = 0; col < count[0].length; col++) {
            for(int row = 1; row < count.length; row++) {
                count[col][row] += count[col][row - 1];
            }
        }

        for (int i = 0; i < count.length; i++) {
            for (int j = 0; j < count[0].length; j++) {
                if ( j < 2 * k && i < 2 * k) continue;
                count[i % (2 * k)][j % (2 * k)] += count[i][j];
            }
        }

        int ans = 0;
        for (int row = 0; row < count.length; row++) {
            for (int col = 0; col < count[0].length; col++) {
                //System.out.printf("%d ", count[col][row]);
                ans = Math.max(ans, count[col][row]);
            }
            //System.out.println();
        }
        System.out.println(ans);
    }
}