Problem:
Score : 500 points

Problem Statement

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.

Sample Input 1

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

Sample Output 1

4
2 3 1 4

There are two vertices directly connected to vertex 2: vertices 3 and 4.
There are also two vertices directly connected to vertex 4: vertices 1 and 2.
Hence, the path 2 → 3 → 1 → 4 satisfies the conditions.

Sample Input 2

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

Sample Output 2

7
1 2 3 4 5 6 7

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>
using namespace __gnu_pbds;

#if !ONLINE_JUDGE
#define debug
#endif

using namespace std;
 
/*******  All Required define Pre-Processors and typedef Constants *******/
#define mem(a, b) memset(a, (b), sizeof(a))
#define repd(i,k) for(int i=k;i>=0;i--)
#define rep(i,k) for(int i=0;i<k;i++)
#define repn(i,k1,k2) for(ll i=k1;i<k2;i++)
#define sz(x) (ll)(x).size()
#define ff first
#define ss second

#define all(cont) cont.begin(), cont.end()
#define rall(cont) cont.end(), cont.begin()
#define mp make_pair
#define pb push_back
#define eb emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define MOD 1000000007
#define ee6 (ll)1000001
#define ee5 (ll)100001

#define trav(a, v) for(auto& a : v)

#define tt ll t;cin>>t; while(t--)
typedef long long int ll;
 
typedef pair<ll, ll> pr;
typedef vector<ll> vi;
typedef vector<string> vs;
typedef vector<pr> vpr;
typedef vector<vpr>vvpr;
typedef vector<vi> vvi;

//*X.find_by_order(2) element at index=2
//X.order_of_key(1) how many elements strictly less than 1

template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag,
    tree_order_statistics_node_update>;

#ifdef debug
#define dbg(...) { cerr<<"[ "; dump(#__VA_ARGS__, __VA_ARGS__);}
#undef endl
template <typename Arg1,typename Arg2>
ostream& operator << (ostream& out, const pair<Arg1,Arg2> &x) {
    return out<<"("<<x.ff<<","<<x.ss<<")";
}
 
template <typename Arg1>
ostream& operator << (ostream& out, const vector<Arg1> &a) {
    out<<"[";for(const auto &x:a)out<<x<<",";return out<<"]";
}
 
template <typename Arg1>
ostream& operator << (ostream& out, const set<Arg1> &a) {
    out<<"[";for(const auto &x:a)out<<x<<",";return out<<"]";
}
 
template <typename Arg1,typename Arg2>
ostream& operator << (ostream& out, const map<Arg1,Arg2> &a) {
    out<<"[";for(const auto &x:a)out<<x<<",";return out<<"]";
}
 
template <typename Arg1,typename Arg2>
ostream& operator << (ostream& out, const unordered_map<Arg1,Arg2> &a) {
    out<<"[";for(const auto &x:a)out<<x<<",";return out<<"]";
} 
 
template <typename Arg1>
void dump(const string name, Arg1&& arg1){
    cerr << name << " : " << arg1 << " ] " << endl;
}
 
template <typename Arg1, typename... Args>
void dump(const string names, Arg1&& arg1, Args&&... args){
    const string name = names.substr(0,names.find(','));
    cerr<<name<<" : "<<arg1<<" | ";
    dump(names.substr(1+(int)name.size()), args...);
}
#else
    #define dbg(args...)
#endif
 
ll powmod(ll x, ll y) 
{ 
    ll res = 1;  
    x = x % MOD;
    while (y > 0) 
    { 
        if (y & 1) 
          res = (res*x) % MOD; 
        y = y>>1; // y = y/2 
        x = (x*x) % MOD;   
    } 
    return res; 
} 
vector<int>g[ee5];
vector<int>ans1,ans2;
bool vis[ee5];
void dfs(int v,vector<int>&ans){
  ans.pb(v);
  vis[v]=true;
  for(auto e:g[v]){
    if(vis[e])continue;    
    dfs(e,ans);
    break;
  }
}

int main()
{
  #if !ONLINE_JUDGE
  //freopen("in.txt","r",stdin);
  #endif
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  
  int n,m;cin>>n>>m;
  rep(i,m){
    int x,y;cin>>x>>y;x--;y--;
    g[x].pb(y);
    g[y].pb(x);
  }   


  dfs(0,ans1);  
  dfs(0,ans2);
  reverse(all(ans2));
  ans2.pop_back();
  cout<<ans2.size()+ans1.size()<<"\n";
  for(auto e:ans2)cout<<e+1<<" ";
  for(auto e:ans1)cout<<e+1<<" ";
}