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.*;
public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int k = sc.nextInt();
		int k2 = 2*k;
		int[][] mat = new int[4*k-1][4*k-1];
		
		for (int i=0;i<n;i++) {
			int x = sc.nextInt();
			int y = sc.nextInt();
			String s = sc.next();
			if (s.equals("W"))
				y += k;
			x %= k2;
			y %= k2;
			
			for (int a=0;a<4;a++) {
				int c = x + a*k;
				if (c>=mat.length)
					break;
				for (int b=0;b<4;b++) {
					int r = y + b*k;
					if (r>=mat.length)
						break;
					int sign = ((a^b)&1)==1 ? -1 : 1;
					mat[c][r] += sign * ((a==0||b==0) ? 1 : 2);
				}
			}
		}
		
		for (int c=0;c<mat.length;c++) {
			for (int r=1;r<mat.length;r++) {
				mat[c][r] += mat[c][r-1];
			}
		}
		for (int r=0;r<mat.length;r++) {
			for (int c=1;c<mat.length;c++) {
				mat[c][r] += mat[c-1][r];
			}
		}

		int ans = 0;
		int x = mat.length-1;
		int y = x;
		for (int c=0;c<k2;c++) {
			for (int r=0;r<k;r++) {
				ans = Math.max(ans, mat[x-c][y-r]);
			}
		}
		
		System.out.println(ans);
	}
}