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

import java.util.*;

public class Main {

	public static void main(String[] args) {
		Main main = new Main();
		main.run();
	}

	public void run() {
		Scanner sc = new Scanner(System.in);

		int n=sc.nextInt();
		int k=sc.nextInt();

		int c[][] = new int[2*k][2*k];
		for(int i=0; i<n; i++) {
			int x=sc.nextInt();
			int y=sc.nextInt();
			if(sc.next().charAt(0)!='B') {
				x+=k;
			}
			x %= 2*k;
			y %= 2*k;
			c[x][y]++;
		}

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

		long max = 0;
		for(int i=0; i<k; i++) {
			for(int j=0; j<k; j++) {
				int tmp1 = s[i][j];
				int tmp2 = s[2*k][j] - s[i+k][j];
				int tmp3 = s[i][2*k] - s[i][j+k];
				int tmp4 = s[i+k][j+k] - s[i][j+k] - s[i+k][j] + s[i][j];
				int tmp5 = s[2*k][2*k] - s[2*k][j+k] - s[i+k][2*k] + s[i+k][j+k];
				int tmp = tmp1+tmp2+tmp3+tmp4+tmp5;
				max = Math.max(max, tmp);
				max = Math.max(max, n-tmp);
			}
		}
		System.out.println(max);
		sc.close();
	}
}
