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.io.BufferedReader;
import java.io.InputStreamReader;

public class Main {
	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String[] sa = br.readLine().split(" ");
		int n = Integer.parseInt(sa[0]);
		int k = Integer.parseInt(sa[1]);
		int k2 = k * 2;
		int k4 = k * 4;
		int[] x = new int[n];
		int[] y = new int[n];
		for (int i = 0; i < n; i++) {
			sa = br.readLine().split(" ");
			int a = Integer.parseInt(sa[0]);
			int b = Integer.parseInt(sa[1]);
			if (sa[2].equals("W")) {
				a += k;
			}
			x[i] = a % k2 + 1;
			y[i] = b % k2 + 1;
		}
		br.close();

		int[] dx = {-k, -k, k, k};
		int[] dy = {-k, k, -k, k};
		int[][] cnt = new int[k4 + 1][k4 + 1];
		for (int i = 0; i < n; i++) {
			cnt[x[i]][y[i]]++;
			for (int j = 0; j < 4; j++) {
				int nx = x[i] + dx[j];
				int ny = y[i] + dy[j];
				if (0 < nx && nx <= k2 && 0 < ny && ny <= k2) {
					cnt[nx][ny]++;
				}
			}
		}
		for (int i = 1; i <= k2; i++) {
			for (int j = 1; j <= k2; j++) {
				cnt[i + k2][j] = cnt[i][j];
				cnt[i][j + k2] = cnt[i][j];
				cnt[i + k2][j + k2] = cnt[i][j];
			}
		}

		for (int i = 0; i <= k4; i++) {
			for (int j = 0; j < k4; j++) {
				cnt[i][j + 1] += cnt[i][j];
			}
		}
		for (int j = 0; j <= k4; j++) {
			for (int i = 0; i < k4; i++) {
				cnt[i + 1][j] += cnt[i][j];
			}
		}

		int ans = 0;
		for (int i = 0; i <= k2; i++) {
			for (int j = 0; j <= k2; j++) {
				int val = cnt[i + k][j + k] - cnt[i][j + k] - cnt[i + k][j] + cnt[i][j];
				ans = Math.max(ans, val);
			}
		}
		System.out.println(ans);
	}
}