Problem: Robot Bender decided to make Fray a birthday present. He drove n nails and numbered them from 1 to n in some order. Bender decided to make a picture using metal rods. The picture is a closed polyline, which vertices should be nails (in the given order). The segments of the polyline should be parallel to the coordinate axes. Polyline is allowed to have self-intersections. Bender can take a rod and fold it exactly once in any place to form an angle of 90 degrees. Then he can attach the place of the fold to some unoccupied nail and attach two ends of this rod to adjacent nails. A nail is considered unoccupied if there is no rod attached to it (neither by it's end nor the by the fold place). No rod could be used twice. It is not required to use all the rods.

Help Bender to solve this difficult task.

Input

The first line contains two positive integers n and m (4 ≤ n ≤ 500, 2 ≤ m ≤ 500, n is even) — the amount of nails and the amount of rods. i-th of the following n lines contains a pair of integers, denoting the coordinates of the i-th nail. Nails should be connected in the same order as they are given in the input. The last line contains m integers — the lenghts of the rods. All coordinates do not exceed 104 by absolute value. Lengths of the rods are between 1 and 200 000. No rod can be used twice. It is guaranteed that all segments of the given polyline are parallel to coordinate axes. No three consecutive nails lie on the same line.

Output

If it is impossible to solve Bender's problem, output NO. Otherwise, output YES in the first line, and in the second line output n numbers — i-th of them should be the number of rod, which fold place is attached to the i-th nail, or -1, if there is no such rod.

If there are multiple solutions, print any of them.

Examples

Input

4 2
0 0
0 2
2 2
2 0
4 4


Output

YES
1 -1 2 -1 


Input

6 3
0 0
1 0
1 1
2 1
2 2
0 2
3 2 3


Output

YES
1 -1 2 -1 3 -1 


Input

6 3
0 0
1 0
1 1
2 1
2 2
0 2
2 2 3


Output

NO

Solution (in JAVA):

Solution:


import java.awt.Point;
import java.io.*;
import java.util.*;
import static java.lang.Math.*;

public class ProblemA_28 {
    
    final boolean ONLINE_JUDGE=System.getProperty("ONLINE_JUDGE")!=null;
    BufferedReader in;
    PrintWriter out;
    StringTokenizer tok=new StringTokenizer("");
    
    void init() throws FileNotFoundException{
        if (ONLINE_JUDGE){
            in=new BufferedReader(new InputStreamReader(System.in));
            out =new PrintWriter(System.out);
        }
        else{
            in = new BufferedReader(new FileReader("input.txt"));
            out = new PrintWriter("output.txt");
        }
    }
    
    String readString() throws IOException{
        while(!tok.hasMoreTokens()){
            tok=new StringTokenizer(in.readLine());
        }
        return tok.nextToken();
    }
    
    int readInt() throws IOException{
        return Integer.parseInt(readString());
    }
    
    public static void main(String[] args){
        new ProblemA_28().run();
    }
    
    public void run(){
        try{
            long t1=System.currentTimeMillis();
            init();
            solve();
            out.close();
            long t2=System.currentTimeMillis();
            System.err.println("Time = "+(t2-t1));
        }catch (Exception e){
            e.printStackTrace(System.err);
            System.exit(-1);
        }
    }
    
    void solve() throws IOException{
        int n = readInt();
        int m = readInt();
        Point[] rod = new Point[n];
        for (int i = 0; i < n; i++){
            rod[i] = new Point(readInt(), readInt());
        }
        Point[] nail = new Point[m];
        for (int i = 0; i < m; i++){
            nail[i] = new Point(readInt(), i);
        }
        Arrays.sort(nail, new Comparator<Point>() {

            @Override
            public int compare(Point p1, Point p2) {
                return p1.x-p2.x;
            }
        });
        Point[] d = new Point[n/2];
        for (int i = 1; i < n-2; i += 2){
            d[i/2] = new Point(d(rod[i], rod[i+1]) + d(rod[i+1], rod[i+2]), i + 1);
        }
        d[d.length-1] = new Point(d(rod[n-1], rod[0]) + d(rod[0], rod[1]), 0);
        boolean[] used = new boolean[m];
        Arrays.fill(used, false);
        int[] uses = new int[n];
        Arrays.fill(uses, -2);
        boolean yes = true;
    a:  for (int i = 0; i < d.length; i++){
            for (int j = 0; j < m; j++){
                if (!used[j] && nail[j].x == d[i].x){
                    uses[d[i].y]=nail[j].y;
                    used[j]=true;
                    continue a;
                }
            }
            yes = false;
            break;
        }
        if (yes){
            out.println("YES");
            for (int i = 0; i < n; i++){
                out.print((uses[i] + 1) + " ");
            }
            return;
        }
        for (int i = 0; i < n-2; i += 2){
            d[i/2] = new Point(d(rod[i], rod[i+1]) + d(rod[i+1], rod[i+2]), i + 1);
        }
        d[d.length-1] = new Point(d(rod[n-2], rod[n-1]) + d(rod[n-1], rod[0]), n-1);
        Arrays.fill(used, false);
        Arrays.fill(uses, -2);
        yes = true;
        a:  for (int i = 0; i < d.length; i++){
            for (int j = 0; j < m; j++){
                if (!used[j] && nail[j].x == d[i].x){
                    uses[d[i].y]=nail[j].y;
                    used[j]=true;
                    continue a;
                }
            }
            yes = false;
            break;
        }
        if (yes){
            out.println("YES");
            for (int i = 0; i < n; i++){
                out.print((uses[i] + 1) + " ");
            }
        }else{
            out.print("NO");
        }
    }       
    
    int d(Point a, Point b){
        return abs(a.x - b.x) + abs(a.y - b.y);
    }
}