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.*;
import java.io.*;

public class Main {
    public static void main(String[] args){
    	FastScanner fs = new FastScanner(System.in);
        int n = fs.nextInt(), k = fs.nextInt();
        int sum[][] = new int[2*k+1][2*k+1];
        int t[][] = new int[2*k][2*k];
        for(int i=0;i<n;i++){
            int x = fs.nextInt(), y = fs.nextInt();
            String c = fs.next();
            x %= (2*k);
            y %= (2*k);
            if(c.equals("B")){
            	x = (x + k) % (2*k);
            }
//            if(x>=k&&y>=k){
//                x-=k;
//                y-=k;
//            }
//            if(y>=k&&x<k){
//                int tmp = x;
//                x = y;
//                y = tmp;
//            }
            t[x][y]++;
        }

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

        int ans = 0;
        for(int i=0;i<=k;i++){
        	for(int j=0;j<=k;j++){
        		int ld = sum[i][j];
        		int rd = sum[2*k][j] - sum[i+k][j];
        		int c = sum[i+k][j+k] - sum[i+k][j] - sum[i][j+k] + sum[i][j];
        		int lu = sum[i][2*k] - sum[i][j+k];
        		int ru = sum[2*k][2*k] - sum[2*k][j+k] - sum[i+k][2*k]  + sum[i+k][j+k];
	            int t_ans =  ld + rd + c + lu + ru;
	            int r_ans = sum[2*k][2*k] - t_ans;
	            ans = Math.max(ans , t_ans);
	            ans = Math.max(ans , r_ans);
        	}
        }
        System.out.println(ans);
    }
}












//高速なScanner
class FastScanner {
	private BufferedReader reader = null;
	private StringTokenizer tokenizer = null;

	public FastScanner(InputStream in) {
		reader = new BufferedReader(new InputStreamReader(in));
		tokenizer = null;
	}

	public String next() {
		if (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				tokenizer = new StringTokenizer(reader.readLine());
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken();
	}

	public String nextLine() {
		if (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				return reader.readLine();
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken("\n");
	}

	public long nextLong() {
		return Long.parseLong(next());
	}

	public int nextInt() {
		return Integer.parseInt(next());
	}

	public double nextDouble() {
		return Double.parseDouble(next());
	}

	public int[] nextIntArray(int n) {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = nextInt();
		return a;
	}

	public long[] nextLongArray(int n) {
		long[] a = new long[n];
		for (int i = 0; i < n; i++)
			a[i] = nextLong();
		return a;
	}
}