Problem: Nudist Beach is planning a military operation to attack the Life Fibers. In this operation, they will attack and capture several cities which are currently under the control of the Life Fibers.

There are n cities, labeled from 1 to n, and m bidirectional roads between them. Currently, there are Life Fibers in every city. In addition, there are k cities that are fortresses of the Life Fibers that cannot be captured under any circumstances. So, the Nudist Beach can capture an arbitrary non-empty subset of cities with no fortresses.

After the operation, Nudist Beach will have to defend the captured cities from counterattack. If they capture a city and it is connected to many Life Fiber controlled cities, it will be easily defeated. So, Nudist Beach would like to capture a set of cities such that for each captured city the ratio of Nudist Beach controlled neighbors among all neighbors of that city is as high as possible. 

More formally, they would like to capture a non-empty set of cities S with no fortresses of Life Fibers. The strength of a city <image> is defined as (number of neighbors of x in S) / (total number of neighbors of x). Here, two cities are called neighbors if they are connnected with a road. The goal is to maximize the strength of the weakest city in S.

Given a description of the graph, and the cities with fortresses, find a non-empty subset that maximizes the strength of the weakest city. 

Input

The first line of input contains three integers n, m, k (2 ≤ n ≤ 100 000, 1 ≤ m ≤ 100 000, 1 ≤ k ≤ n - 1).

The second line of input contains k integers, representing the cities with fortresses. These cities will all be distinct. 

The next m lines contain the roads. The i-th of these lines will have 2 integers ai, bi (1 ≤ ai, bi ≤ n, ai ≠ bi). Every city will have at least one road adjacent to it.

There is no more than one road between each pair of the cities.

Output

The first line should contain an integer r, denoting the size of an optimum set (1 ≤ r ≤ n - k). 

The second line should contain r integers, denoting the cities in the set. Cities may follow in an arbitrary order. This line should not contain any of the cities with fortresses.

If there are multiple possible answers, print any of them.

Examples

Input

9 8 4
3 9 6 8
1 2
1 3
1 4
1 5
2 6
2 7
2 8
2 9


Output

3
1 4 5


Input

10 8 2
2 9
1 3
2 9
4 5
5 6
6 7
7 8
8 10
10 4


Output

8
1 5 4 8 10 6 3 7

Note

The first example case achieves a strength of 1/2. No other subset is strictly better.

The second example case achieves a strength of 1. Note that the subset doesn't necessarily have to be connected.

Solution (in JAVA):

Solution:

import java.io.*;
import java.lang.reflect.Array;
import java.math.*;
import java.text.DecimalFormat;
import java.util.*;




public class Main {
    private static ArrayList<Integer> [] G;
    private static int n,m,k;
    private static boolean [] dont,dead;
    private static int [] cnt;

    public static void main(String[] args) throws Exception{
        IO io = new IO(null,null);
        n = io.getNextInt();
        m = io.getNextInt();
        k = io.getNextInt();
        dont = new boolean[n+1];
        dead = new boolean[n+1];
        G = new ArrayList[n+1];
        cnt = new int[n+1];

        for (int i = 1;i <= n;i++)
            G[i] = new ArrayList<>();

        for (int i = 0;i < k;i++) {
            int x = io.getNextInt();
            dead[x] = dont[x] = true;
        }
        for (int e = 0;e < m;e++) {
            int a = io.getNextInt();
            int b = io.getNextInt();
            G[a].add(b);
            G[b].add(a);
        }

        TreeSet<node> TS = new TreeSet<>();

        for (int i = 1;i <= n;i++)
            if (!dont[i]) {
                for (int x : G[i])
                    if (!dont[x])
                        cnt[i]++;
                TS.add(new node(cnt[i],G[i].size(),i));
            }
        node best = new node(0,1,1);
        while (!TS.isEmpty()) {
            node U = TS.pollFirst();
            if (U.compareTo(best) > 0) best = U;
            dead[U.u] = true;
            for (int v : G[U.u]) if (!dead[v]){
                TS.remove(new node(cnt[v],G[v].size(),v));
                cnt[v]--;
                TS.add(new node(cnt[v],G[v].size(),v));
            }
        }

//        System.err.println(best.a + "/" + best.b);

        for (int i = 1;i <= n;i++) dead[i] = dont[i];
        for (int i = 1;i <= n;i++)
            if (!dont[i]) {
                cnt[i] = 0;
                for (int x : G[i])
                    if (!dont[x])
                        cnt[i]++;
                TS.add(new node(cnt[i],G[i].size(),i));
            }

        while (!TS.isEmpty()) {
            node U = TS.first();
            if (U.compareTo(best) == 0) break;
            TS.pollFirst();
            dead[U.u] = true;
            for (int v : G[U.u]) if (!dead[v]){
                TS.remove(new node(cnt[v],G[v].size(),v));
                cnt[v]--;
                TS.add(new node(cnt[v],G[v].size(),v));
            }
        }

        io.println(TS.size());
        for (node U : TS) io.print(U.u + " ");
        io.println("");
        io.close();
    }
}


class node implements Comparable<node> {
    int a,b, u;

    public node(int a,int b,int u) {
        this.a = a;
        this.b = b;
        this.u = u;
    }

    @Override
    public int compareTo(node o) {
        long diff = a*(long)o.b - b*(long)o.a;
        if (diff < 0) return -1;
        if (diff > 0) return 1;
        return this.u - o.u;
    }
}



class IO{
    private BufferedReader br;
    private StringTokenizer st;
    private PrintWriter writer;
    private String inputFile,outputFile;

    public boolean hasMore() throws IOException{
        if(st != null && st.hasMoreTokens()) return true;
        if(br != null && br.ready()) return true;
        return false;
    }
    public String getNext() throws FileNotFoundException, IOException{
        while(st == null || !st.hasMoreTokens()) st = new StringTokenizer(br.readLine());
        return st.nextToken();
    }

    public String getNextLine() throws FileNotFoundException, IOException{
        return br.readLine().trim();
    }

    public int getNextInt() throws FileNotFoundException, IOException{
        return Integer.parseInt(getNext());
    }
    public long getNextLong() throws FileNotFoundException, IOException{
        return Long.parseLong(getNext());
    }

    public void print(double x,int num_digits) throws  IOException{
        writer.printf("%." + num_digits + "f" ,x);
    }
    public void println(double x,int num_digits) throws  IOException{
        writer.printf("%." + num_digits + "f\n" ,x);
    }
    public void print(Object o) throws  IOException{
        writer.print(o.toString());
    }

    public void println(Object o) throws  IOException{
        writer.println(o.toString());
    }
    public IO(String x,String y) throws FileNotFoundException, IOException{
        inputFile = x;
        outputFile = y;
        if(x != null) br = new BufferedReader(new FileReader(inputFile));
        else br = new BufferedReader(new InputStreamReader(System.in));
        if(y != null) writer = new PrintWriter(new BufferedWriter(new FileWriter(outputFile)));
        else writer = new PrintWriter(new OutputStreamWriter(System.out));
    }

    protected void close() throws IOException{
        br.close();
        writer.close();
    }
    public void outputArr(Object [] A) throws IOException{
        int L = A.length;
        for (int i = 0;i < L;i++) {
            if(i > 0) writer.print(" ");
            writer.print(A[i]);
        }
        writer.print("\n");
    }
}