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.Scanner;
 
public class Main {
	public static void main(String[] args) {
		final Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int k = sc.nextInt();
		int k2 = k * 2;
		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();
			if (sc.next().equals("W")) {
				y += k;//全てblackに
			}
			x %= k2;
			y %= k2;
			//二次元imos 重みの加算
			count[x][y]++;
			count[x][y + k]--;
			count[x + k][y]--;
			count[x + k][y + k]++;
			count[x + k][y + k]++;
			count[x + k][y + k + k]--;
			count[x + k + k][y + k]--;
			count[x + k + k][y + k + k]++;
		}
		//横方向への累積和
		final int H = count.length;
		final int W = count[0].length;
		for (int y = 0; y < H; y++) {
			for (int x = 1; x < W; x++) {
				count[y][x] += count[y][x - 1];
			}
		}
		//縦方向への累積和
		for (int y = 1; y < H; y++) {
			for (int x = 0; x < W; x++) {
				count[y][x] += count[y - 1][x];
			}
		}
		for(int y = 0; y < H; y++) {
			for(int x = 0; x < W; x++) {
				if(y < k2 && x < k2) continue;
				count[y % k2][x % k2] += count[y][x];
			}
		}
		long ans = 0L;
		for (int y = 0; y < k2; y++) {
			for (int x = 0; x < k2; x++) {
				ans = Math.max(ans, count[y][x]);
			}
		}
		System.out.println(ans);
	}
}