Problem: You are given a connected undirected simple graph, which has N vertices and M edges. The vertices are numbered 1 through N, and the edges are numbered 1 through M. Edge i connects vertices A_i and B_i. Your task is to find a path that satisfies the following conditions:

* The path traverses two or more vertices.
* The path does not traverse the same vertex more than once.
* A vertex directly connected to at least one of the endpoints of the path, is always contained in the path.



It can be proved that such a path always exists. Also, if there are more than one solution, any of them will be accepted.

Constraints

* 2 \leq N \leq 10^5
* 1 \leq M \leq 10^5
* 1 \leq A_i < B_i \leq N
* The given graph is connected and simple (that is, for every pair of vertices, there is at most one edge that directly connects them).

Input

Input is given from Standard Input in the following format:


N M
A_1 B_1
A_2 B_2
:
A_M B_M


Output

Find one path that satisfies the conditions, and print it in the following format. In the first line, print the count of the vertices contained in the path. In the second line, print a space-separated list of the indices of the vertices, in order of appearance in the path.

Examples

Input

5 6
1 3
1 4
2 3
1 5
3 5
2 4


Output

4
2 3 1 4


Input

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


Output

7
1 2 3 4 5 6 7

Solution (in JAVA):

Solution:
import java.util.*;import java.io.*;import java.math.*;
public class Main{
  //↓見なくていいよ！ここから------------------------------------------
  static class InputIterator{
    ArrayList<String> inputLine = new ArrayList<String>(1024);
    int index = 0; int max;
    InputIterator(){
      BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
      while(true){
        String read;
        try{
          read = br.readLine();
        }catch(IOException e){
          read = null;
        }
        if(read != null){
          inputLine.add(read);
        }else{
          break;
        }
      }
      max = inputLine.size();
    }
    boolean hasNext(){return (index < max);}
    String next(){
      if(hasNext()){
        String returnStr = inputLine.get(index);
        index++;
        return returnStr;
      }else{
        throw new IndexOutOfBoundsException("これ以上入力はないよ");
      }
    }
  }
  
  static InputIterator ii = new InputIterator();//リアクティブでは使えないので諦めてScanner使うこと
  static PrintWriter out = new PrintWriter(System.out);
  static void flush(){out.flush();}
  static void myout(Object t){out.println(t);}
  static void myerr(Object t){System.err.print("debug:");System.err.println(t);}
  static String next(){return ii.next();}
  static int nextInt(){return Integer.parseInt(next());}
  static long nextLong(){return Long.parseLong(next());}
  static ArrayList<String> nextStrArray(){return myHanSpSplit(next());}
  static ArrayList<String> myHanSpSplit(String str){return new ArrayList<String>(Arrays.asList(str.split(" ")));}
  static ArrayList<String> nextCharArray(){return mySingleSplit(next());}
  static ArrayList<String> mySingleSplit(String str){return new ArrayList<String>(Arrays.asList(str.split("")));}
  static ArrayList<Integer> nextIntArray(){
    ArrayList<Integer> ret = new ArrayList<Integer>();
    ArrayList<String> input = nextStrArray();
    for(int i = 0; i < input.size(); i++){
      ret.add(Integer.parseInt(input.get(i)));
    }
    return ret;
  }
  static ArrayList<Long> nextLongArray(){
    ArrayList<Long> ret = new ArrayList<Long>();
    ArrayList<String> input = nextStrArray();
    for(int i = 0; i < input.size(); i++){
      ret.add(Long.parseLong(input.get(i)));
    }
    return ret;
  }
  static String kaigyoToStr(String[] list){return String.join("\n",list);}
  static String kaigyoToStr(ArrayList<String> list){return String.join("\n",list);}
  static String hanSpToStr(String[] list){return String.join(" ",list);}
  static String hanSpToStr(ArrayList<String> list){return String.join(" ",list);}
  public static void main(String[] args){
    Runtime rt = Runtime.getRuntime();
    long mae = System.currentTimeMillis();
    solve();
    flush();
    long ato = System.currentTimeMillis();
    myerr("time : " + (ato - mae) + "ms");
    myerr("memory : " + ((rt.totalMemory() - rt.freeMemory()) / 1024) + "KB");
  }
  //↑見なくていいよ！ここまで------------------------------------------
static Node[] list;
static ArrayDeque<Integer> output;
  static void solve(){//ここがメイン関数代わり
	ArrayList<Integer> one = nextIntArray();
	int N = one.get(0);
	int M = one.get(1);
	list = new Node[N];
	for(int i = 0; i < N; i++){
		list[i] = new Node(i);
	}
	for(int i = 0; i < M; i++){
		ArrayList<Integer> tmp = nextIntArray();
		list[tmp.get(0) - 1].child.add(tmp.get(1) - 1);
		list[tmp.get(1) - 1].child.add(tmp.get(0) - 1);
	}
	output = new ArrayDeque<Integer>();
	list[0].access = true;
	output.addFirst(1);
	dfs(0, true);
	dfs(0, false);
	myout(output.size());
	ArrayList<String> tmpOut = new ArrayList<String>();
	while(output.size() > 0){
		tmpOut.add(String.valueOf(output.pollFirst()));
	}
	myout(hanSpToStr(tmpOut));
  }
  //Method addition frame start
public static void dfs(int pushIndex, boolean isBack){
	if(isBack){
		if(pushIndex != 0){
			output.addLast(pushIndex + 1);
		}
	}else{
		if(pushIndex != 0){
			output.addFirst(pushIndex + 1);
		}
	}
	ArrayList<Integer> child = list[pushIndex].child;
	for(int i = 0; i < child.size(); i++){
		if(!list[child.get(i)].access){
			list[child.get(i)].access = true;
			dfs(child.get(i), isBack);
			break;
		}
	}
}
public static class Node{
	int no;
	ArrayList<Integer> child = new ArrayList<Integer>();
	boolean access = false;
	Node(int no){
		this.no = no;
	}
}
  //Method addition frame end
}