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: Wrong Answer)

#include <iostream>
#include <string>
#include <map>
#include <algorithm>
#include <list>
#include <vector>
#include <set>
#include <stdio.h>
#include <queue>
#include <stack>
#include <deque>
#include <math.h>
#include <sstream>
#include <stdlib.h>
#include <functional>
using namespace std;
 
#define rep(i,n) for(int i = 0; i < n; i++)
#define rrep(i,a,n) for(int i = a; i < n; i++)
#define INF (1<<29)
#define INFL 0x3f3f3f3f3f3f3f3fLL
#define MOD 1000000007
#define fi first
#define se second 
#define pb push_back
#define PI 3.14159265358979323846
#define all(o) (o).begin(), (o).end()
#define rall(x) x.rbegin(),x.rend()
 
 
typedef double ld;
typedef vector<int> vi;
typedef vector< vi > vvi;
typedef vector<string> vs;
typedef vector<char> vc;
typedef vector<ld> vd;
typedef vector < vc > vvc;
typedef long long ll;
typedef pair<int, int> pii;
typedef unsigned long long int ulli;
 
const int dx[] = { 1, 0, -1, 0 };
const int dy[] = { 0, 1, 0, -1 };
 
template <typename T> T &chmin(T &a, const T &b) { return a = min(a, b); }
template <typename T> T &chmax(T &a, const T &b) { return a = max(a, b); }
 
const ld eps = 1e-10, pi = acos(-1.0);
 
#define MAX_N 100001
 
// combination
 
ll fact[MAX_N], factinv[MAX_N];
ll mod_pow(ll n, ll p, ll m) {
	ll a = n;
	ll x = 1;
	while (p) {
		if (p & 1) x = (x * a) % m;
		a = (a * a) % m;
		p >>= 1;
	}
	return x;
}
int extgcd(int a, int b, int& x, int&y) {
	int d = a;
	if (b != 0) {
		d = extgcd(b, a % b, y, x);
		y -= (a / b) * x;
	}
	else {
		x = 1; y = 0;
	}
	return d;
}
int mod_inverse(int a, int m) {
	int x, y;
	extgcd(a, m, x, y);
	return (m + x % m) % m;
}
ll inv(ll n) {
	return mod_pow(n, MOD - 2, MOD);
}
void combInit() {
	fact[0] = 1;
	rrep(i, 1, MAX_N) fact[i] = fact[i - 1] * i % MOD;
	factinv[MAX_N - 1] = inv(fact[MAX_N - 1]);
	for (int i = MAX_N - 2; i >= 0; i--) factinv[i] = factinv[i + 1] * (i + 1) % MOD;
}
 
// ダイクストラ
/*
struct edge { int to, cost; };
vector < vector < edge > > G;
vi d;
void dijkstra(int s) {priority_queue<pii, vector< pii >, greater<pii> > que;d[s] = 0;que.push(pii(0, s));while (!que.empty()) {pii p = que.top(); que.pop();int v = p.second;if (d[v] < p.first) continue;rep(i, G[v].size()) {edge e = G[v][i];int cost = e.cost;if (d[v] + cost < d[e.to]) {d[e.to] = d[v] + e.cost;que.push(pii(d[e.to], e.to));}}}}
*/
 
// Union-Find木
int par[MAX_N], rnk[MAX_N], unionSize[MAX_N];
// はじめは全ての頂点が根
void UnionFindTreeInit(int n) {
	rep(i, n) {
		par[i] = i;
		rnk[i] = 0;
		unionSize[i] = 1;
	}
}
//木の根元を求める
int root(int x) {
	if (par[x] == x) return x; // 根を返す
	else return par[x] = root(par[x]);
}
// 連結成分の大きさを取得
int componentSize(int x) {
	return unionSize[root(x)];
}
// xとyが同じ集合に属するか否か
bool same(int x, int y) {
	return root(x) == root(y);
}
// xとyの属する集合を併合
void unite(int x, int y) {
	x = root(x);
	y = root(y);
	if (x == y) return;
	if (rnk[x] < rnk[y]) {
		par[x] = y;
		unionSize[y] += unionSize[x];
	}
	else {
		par[y] = x;
		if (rnk[x] == rnk[y]) rnk[x]++;
		unionSize[x] += unionSize[y];
	}
}
 
 
//-----------------------------------------------------------------
 
//int dx[8] = { -1, 1, 0, 0, 1, 1, -1, -1 };
//int dy[8] = { 0, 0, 1, -1, 1, -1, -1, 1 };
/* 
struct edge { int to, cost; };
vector < vector < edge > > G;
vi d;
void dijkstra(int s) {
	priority_queue<pii, vector< pii >, greater<pii> > que;
	d[s] = 0;
	que.push(pii(0, s));
 
	while (!que.empty()) {
		pii p = que.top();
		que.pop();
		int v = p.second;
		if (d[v] < p.first) continue;
 
		rep(i, G[v].size()) {
			edge e = G[v][i];
			int cost = e.cost;
			if (d[v] + cost < d[e.to]) {
				d[e.to] = d[v] + e.cost;
				que.push(pii(d[e.to], e.to));
			}
		}
	}
 
}
*/
//-------------------------------------------------------

int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	ll n,m;
	cin >> n >> m;
	vector <vector<ll> > G(n);
	rep(i,m) {
		ll a,b;
		cin >> a >> b;
		a--; b--;
		G[a].push_back(b);
		G[b].push_back(a);
	}
	rep(i,n) {
		sort(all(G[i]));
	}
	UnionFindTreeInit(n);
	vi a;
	a.push_back(0);
	ll p = 0;
	while(G[p].size()) {

		if (!same(p, G[p][0])) {
			unite(p,G[p][0]);
			ll t = G[p][0];
			G[p].erase(G[p].begin());
			p = t;
			a.push_back(p);	
			continue;
		}
		G[p].erase(G[p].begin());
	}
	cout << a.size() <<endl;
	rep(i, a.size()) {
		cout << a[i] + 1 << " ";
	}
	cout << endl;
	return 0;
}