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();
        final int[][] table = new int[2 * k][2 * k];

        for (int i = 0; i < n; i++) {
            int x = scanner.nextInt() % (2 * k);
            final int y = scanner.nextInt() % (2 * k);
            final String color = scanner.next();
            if ("B".equals(color)) {
                x = (x + k) % (2 * k);
            }

            table[x][y]++;
        }

        for (int i = 1; i < 2 * k; i++) {
            for (int j = 0; j < 2 * k; j++) {
                table[i][j] += table[i - 1][j];
            }
        }

        for (int i = 0; i < 2 * k; i++) {
            for (int j = 1; j < 2 * k; j++) {
                table[i][j] += table[i][j - 1];
            }
        }

        int max = 0;
        for (int i = 0; i < k; i++) {
            for (int j = 0; j < k; j++) {
                int sum;
                if (i == 0 & j == 0) {
                    sum = table[k - 1][k - 1]
                            + table[2 * k - 1][2 * k - 1] - (table[2 * k - 1][k - 1] + table[k - 1][2 * k - 1] - table[k - 1][k - 1]);
                } else if (i == 0) {
                    sum = table[k - 1][j + k - 1] - table[k - 1][j - 1]
                            + table[2 * k - 1][j - 1] - table[k - 1][j - 1]
                            + table[2 * k - 1][2 * k - 1] - (table[2 * k - 1][j + k - 1] + table[k - 1][2 * k - 1] - table[k - 1][j + k - 1]);
                } else if (j == 0) {
                    sum = table[i + k - 1][k - 1] - table[i - 1][k - 1]
                            + table[i - 1][2 * k - 1] - table[i - 1][k - 1]
                            + table[2 * k - 1][2 * k - 1] - (table[i + k - 1][2 * k - 1] + table[2 * k - 1][k - 1] - table[i + k - 1][k - 1]);
                } else {
                    sum = table[i + k - 1][j + k - 1] - (table[i + k - 1][j - 1] + table[i - 1][j + k - 1] - table[i - 1][j - 1]) // center
                            + table[2 * k - 1][j - 1] - table[i + k - 1][j - 1] //left top
                            + table[i - 1][j - 1] // left bottom
                            + table[2 * k - 1][2 * k - 1] - (table[2 * k - 1][j + k - 1] + table[i + k - 1][2 * k - 1] - table[i + k - 1][j + k - 1]) // right top
                            + table[i - 1][2 * k - 1] - table[i - 1][j + k - 1]; // right down
                }
                max = Math.max(max, Math.max(sum, n - sum));
            }
        }

        System.out.println(max);
    }

}