Problem: You are given an undirected connected graph consisting of n vertices and m edges. Your goal is to destroy all edges of the given graph.

You may choose any vertex as the starting one and begin walking from it along the edges. When you walk along an edge, you destroy it. Obviously, you cannot walk along an edge if it is destroyed.

You can perform the mode shift operation at most once during your walk, and this operation can only be performed when you are at some vertex (you cannot perform it while traversing an edge). After the mode shift, the edges you go through are deleted in the following way: the first edge after the mode shift is not destroyed, the second one is destroyed, the third one is not destroyed, the fourth one is destroyed, and so on. You cannot switch back to the original mode, and you don't have to perform this operation if you don't want to.

Can you destroy all the edges of the given graph?

Input

The first line contains two integers n and m (2 ≤ n ≤ 3000; n - 1 ≤ m ≤ min((n(n-1))/(2), 3000)) — the numbef of vertices and the number of edges in the graph.

Then m lines follow, each containing two integers x_i and y_i (1 ≤ x_i, y_i ≤ n; x_i ≠ y_i) — the endpoints of the i-th edge. 

These edges form a connected undirected graph without multiple edges.

Output

If it's impossible to destroy all of the edges, print 0.

Otherwise, print the sequence of your actions as follows. First, print k — the number of actions (k ≤ 2m + 2). Then, print the sequence itself, consisting of k integers. The first integer should be the index of the starting vertex. Then, each of the next integers should be either the index of the next vertex in your traversal, or -1 if you use mode shift. You are allowed to use mode shift at most once.

If there are multiple answers, print any of them.

Examples

Input


3 3
1 2
2 3
3 1


Output


4
1 2 3 1


Input


4 3
1 2
2 3
4 2


Output


8
2 -1 1 2 3 2 4 2 


Input


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


Output


9
2 3 1 2 -1 4 2 5 2 


Input


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


Output


8
5 4 2 3 1 -1 2 1 


Input


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


Output


0

Solution (in CPP):

Solution:
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <vector>
#include <bitset>
#include <string>
#include <cstring>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <deque>
#include <utility>
#include <algorithm>
#include <random>
#include <cmath>
#include <cassert>
#include <climits>
#include <ctime>
#include <chrono>



#pragma GCC optimize("Ofast")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,sse4.1,sse4.2,popcnt,abm,mmx,avx,avx2,tune=native")



#ifdef LOCAL
    #define dbg(x) cout << #x << " : " << x << endl;
#else
    #define dbg(x)
#endif

#define pb push_back
#define ppb pop_back()
#define mp make_pair
#define fi(a, b) for (int i = a; i < b; i++)
#define fj(a, b) for (int j = a; j < b; j++)
#define fk(a, b) for (int k = a; k < b; k++)
#define fi1(a, b) for (int i = a - 1; i >= b; i--)
#define fj1(a, b) for (int j = a - 1; j >= b; j--)
#define fk1(a, b) for (int k = a - 1; k >= b; k--)
#define fx(x, a) for (auto& x : a)
#define rep(i, a, b) for (int i = a; i < b; ++i)
#define rep1(i, a, b) for (int i = a - 1; i >= b; --i)
#define siz(x) (int)x.size()
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end()
#define left leftkekmda
#define right rightkekmda

using namespace std;

template<typename T1, typename T2>inline void mine(T1 &x, const T2 &y) { if (y < x) x = y; }
template<typename T1, typename T2>inline void maxe(T1 &x, const T2 &y) { if (x < y) x = y; }

template<typename T> ostream& operator << (ostream &os, const vector<T> &b) {
    for (auto &k : b) os << k << ' ';
    return os;
}
template<typename T> istream& operator >> (istream &is, vector<T> &b) {
    for (auto &k : b) is >> k;
    return is;
}


typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef char ch;
typedef string str;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<pii> vpii;
typedef vector<vpii> vvpii;
typedef vector<ch> vch;
typedef vector<vch> vvch;
typedef vector<str> vs;


mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());



const int N = 3010;
int n, m;
namespace dsu {
    int p[N];
    int sz[N];
    int eds[N];
    int cnt;
    struct inf {
        int a;
        int b;
        int pra;
        int prb;
        int sza;
        int szb;
        int ea;
        int eb;
        int pcnt;
        inf (int _a, int _b, int _pcnt) {
            a = _a;
            ea = eds[a];
            pra = p[a];
            sza = sz[a];
            b = _b;
            eb = eds[b];
            prb = p[b];
            szb = sz[b];
            pcnt = _pcnt;
        }
    };
    vector<inf> pers;
    void init() {
        for (int i = 0; i < n; ++i) {
            p[i] = i;
            sz[i] = 1;
            eds[i] = 0;
        }
        cnt = 0;
    }
    int get(int v) {
        if (p[v] == v)
            return v;
        return get(p[v]);
    }
    void unite(int a, int b) {
        //dbg(cnt);
        //cout << "unite: " << a << ' ' << b << '\n';
        a = get(a);
        b = get(b);
        pers.pb(inf(a, b, cnt));
        if (a == b) {
            cnt += eds[a] == 0;
            eds[a]++;
            return;
        }
        if (sz[a] < sz[b]) swap(a, b);
        cnt -= eds[a] > 0;
        cnt -= eds[b] > 0;
        p[b] = a;
        sz[a] += sz[b];
        eds[a] += eds[b] + 1;
        ++cnt;
    }
    void rollback() {
        inf i = pers.back();
        //cout << "split: " << i.a << ' ' << i.b << '\n';
        sz[i.a] = i.sza;
        p[i.a] = i.pra;
        eds[i.a] = i.ea;
        sz[i.b] = i.szb;
        p[i.b] = i.prb;
        eds[i.b] = i.eb;
        
        cnt = i.pcnt;
        //dbg(cnt);
        pers.ppb;
    }
}

vpii dcp[4 * N];

void add(int v, int l, int r, pii edge, int cl, int cr) {
    if (cl >= r || cr <= l)
        return;
    if (cl >= l && cr <= r) {
        dcp[v].pb(edge);
        return;
    }
    int mid = (cl + cr) / 2;
    add(2 * v, l, r, edge, cl, mid);
    add(2 * v + 1, l, r, edge, mid, cr);
}

vi g[N];
bool c[N][N];
vi odd[N];
vi eul;
void dfs(int v, int start, const vch & ban) {
    while (g[v].size()) {
        int to = g[v].back();
        g[v].ppb;
        if (!c[v][to] || (v == start && ban[to]) || (ban[v] && to == start)) continue;
        c[v][to] = c[to][v] = 0;
        dfs(to, start, ban);
    }
    eul.pb(v);
}

void check(int v) {
    int r = -1;
    for (int i = 0; i < n; ++i) {
        if (i != v && !c[v][i] && g[i].size() % 2) {
            if (r != -1)
                return;
            r = i;
        }
    }
    //dbg(dsu::cnt);
    auto con = [&]() {
        return dsu::cnt == 0 || (dsu::cnt == 1 && dsu::eds[dsu::get(v)] > 0);
    };
    //dbg(v);
    //dbg(r);
    //dbg(odd[v]);
    int skip = -1;
    vch ban(n, 0);
    fx(to, odd[v]) ban[to] = 1;
    if (r != -1) {
        if (!con()) return;
        dfs(v, v, ban);
    } else {
        if (con())
            dfs(v, v, ban);
        fx(to, odd[v]) {
            if (eul.size()) break;
            dsu::unite(v, to);
            if (con()) {
                ban[to] = false;
                dfs(v, v, ban);
                skip = to;
            }
            dsu::rollback();
        }
        if (eul.size() == 0) return;
    }
    assert(eul.size());
    eul.pb(-2);
    fx(to, odd[v]) {
        if (to != skip) {
            eul.pb(to);
            eul.pb(v);
        }
    }
    cout << eul.size() << '\n';
    //cout << eul << '\n';
    fx(x, eul) {
        cout << x + 1 << ' ';
    }
    cout << '\n';
    exit(0);
}

void solve(int v, int l, int r) {
    for (auto[a, b] : dcp[v])
        dsu::unite(a, b);
    if (l + 1 == r) {
        //dbg(l);
        check(l);
    } else {
        int mid = (l + r) / 2;
        solve(2 * v, l, mid);
        solve(2 * v + 1, mid, r);
    }
    fx(x, dcp[v])
        dsu::rollback();
}


int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    cin >> n >> m;
    dsu::init();
    while (m--) {
        int a, b;
        cin >> a >> b;
        --a, --b;
        g[a].pb(b);
        g[b].pb(a);
        c[a][b] = c[b][a] = 1;
    }
    for (int i = 0; i < n; ++i) {
        for (int j = i + 1; j < n; ++j) {
            if (!c[i][j]) continue;
            add(1, 0, i, {i, j}, 0, n);
            add(1, i + 1, j, {i, j}, 0, n);
            add(1, j + 1, n, {i, j}, 0, n);
            if (g[j].size() % 2 == 0)
                add(1, i, i + 1, {i, j}, 0, n);
            else 
                odd[i].pb(j);

            if (g[i].size() % 2 == 0)
                add(1, j, j + 1, {i, j}, 0, n);
            else
                odd[j].pb(i);
        }
    }
    solve(1, 0, n);
    cout << "0\n";
}