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)

#pragma GCC optimize ("O3")
#pragma GCC target ("tune=native")
#pragma GCC target ("avx")

#include <bits/stdc++.h>

// 汎用マクロ
#define ALL_OF(x) (x).begin(), (x).end()
#define REP(i,n) for (long long i=0, i##_len=(n); i<i##_len; i++)
#define RANGE(i,is,ie) for (long long i=(is), i##_end=(ie); i<=i##_end; i++)
#define DSRNG(i,is,ie) for (long long i=(is), i##_end=(ie); i>=i##_end; i--)
#define UNIQUE(v) { sort((v).begin(), (v).end()); (v).erase(unique((v).begin(), (v).end()), (v).end()); }
template<class T> bool chmax(T &a, const T &b) {if (a < b) {a = b; return 1;} return 0; }
template<class T> bool chmin(T &a, const T &b) {if (a > b) {a = b; return 1;} return 0; }
#define INF 0x7FFFFFFF
#define LINF 0x7FFFFFFFFFFFFFFFLL
#define Yes(q) (q ? "Yes" : "No")
#define YES(q) (q ? "YES" : "NO")
#define DUMP(q) cerr << "[DEBUG] " #q ": " << (q) << " at " __FILE__ ":" << __LINE__ << endl
#define DUMPALL(q) { cerr << "[DEBUG] " #q ": ["; REP(dumpall_i, (q).size()) { cerr << q[dumpall_i] << (dumpall_i == (q).size() - 1 ? "" : ", "); } cerr << "] at " __FILE__ ":" << __LINE__ << endl; }

// gcc拡張マクロ
#define gcd __gcd
#define popcount __builtin_popcount
#define popcountll __builtin_popcountll

// エイリアス
using  ll = long long;
using ull = unsigned long long;
using  ld = long double;
using namespace std;

// モジュール


// 処理内容
int main() {
    
    ll n, m; cin >> n >> m;
    vector<vector<ll>> nbr(n);
    REP(i, m) {
        ll a, b; cin >> a >> b;
        --a; --b; // 0-indexed
        nbr[a].push_back(b);
        nbr[b].push_back(a);
    }

    deque<ll> ans = {0, nbr[0][0]};
    vector<bool> visited(n, false);
    visited[ans[0]] = visited[ans[1]] = true;
    for (;;) {
        ll v = ans.back();
        bool yet = false;
        for (ll nb : nbr[v]) {
            if (!visited[nb]) {
                ans.push_back(nb);
                visited[nb] = true;
                yet = true;
                break;
            }
        }
        if (!yet) break;
    }
    for (;;) {
        ll v = ans.front();
        bool yet = false;
        for (ll nb : nbr[v]) {
            if (!visited[nb]) {
                ans.push_front(nb);
                visited[nb] = true;
                yet = true;
                break;
            }
        }
        if (!yet) break;
    }

    cout << ans.size() << endl;
    for (ll &v : ans)
        cout << -~v << " \n"[&v == &ans.back()];
    
}