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;
import java.util.stream.IntStream;

/**
 * 解説どおりに作成したソースコード<br/>
 * https://atcoder.jp/contests/abc086/submissions/2279952 にも参照
 */
public class Main {

	/** 白の定数 */
	private static final String WHITE = "W";

	/** 四角の差分を計算用 */
	private static final int[] X_DIFFS = { 0, 0, 1, 1, 1, 1, 2, 2 };
	private static final int[] Y_DIFFS = { 0, 1, 0, 1, 1, 2, 1, 2 };
	private static final int[] COUNT_DIFFS = { 1, -1, -1, 1, 1, -1, -1, 1 };

	public static void main(String[] args) {
		try (Scanner scanner = new Scanner(System.in)) {
			int n = scanner.nextInt();
			int k = scanner.nextInt();
			int w = 4 * k + 1;
			int k2 = 2 * k;
			int[][] count = new int[w][w];
			IntStream.range(0, n).forEach(i -> {
				int x = scanner.nextInt();
				int y = scanner.nextInt();
				if (WHITE.equals(scanner.next())) {
					y += k;
				}
				x %= 2 * k;
				y %= 2 * k;
				for (int j = 0; j < X_DIFFS.length; j++) {
					count[x + X_DIFFS[j] * k][y + Y_DIFFS[j] * k] += COUNT_DIFFS[j];
				}
			});
			// 横方向への累積和
			IntStream.range(0, w).forEach(i -> IntStream.range(1, w).forEach(j -> count[i][j] += count[i][j - 1]));
			// 縦方向への累積和
			IntStream.range(1, w).forEach(i -> IntStream.range(0, w).forEach(j -> count[i][j] += count[i - 1][j]));
			// 2k x 2k 範囲を超えたcountを2k x 2k範囲に加算
			IntStream.range(0, w).forEach(i -> IntStream.range(0, w).filter(j -> (i >= k2) || (j >= k2))
					.forEach(j -> count[i % k2][j % k2] += count[i][j]));
			// 2k x 2k 範囲の最大値を出力
			System.out.println(IntStream.range(0, k2)
					.map(i -> IntStream.range(0, k2).map(j -> count[i][j]).max().getAsInt()).max().getAsInt());
		}
	}
}