// #pragma GCC optimize(2)
// #pragma GCC optimize(3,"Ofrrst","inline")
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl "\n"
#define ull unsigned long long
#define ll __int128_t
#define fre freopen("input.txt", "r", stdin);freopen("output.txt", "w", stdout);
#define ios ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
mt19937 rd(chrono::system_clock::now().time_since_epoch().count());
const int mod=998244353;
const int inf=0x3f3f3f3f3f3f3f3f;
const int N=2e5+10;
const double eps=1e-6;
const int hash_p1=1610612741;
const int hash_p2=805306457;
const int hash_p3=402653189;
const int base_1=131;
const int base_2=13331;
int n,m;
#define pre(i,a,b) for(int i=a;i<=b;i++)
vector<int> g[N];
int dfn[N],low[N],deep,cut[N],cnt;
// v:当前点 r：本次搜索树的root
void tarjan(ll u, ll r) {
	dfn[u] = low[u] = ++deep;
	ll child = 0;
	for (auto v : g[u]) {
		if (!dfn[v]) {
			tarjan(v, r);
			low[u] = min(low[u], low[v]);
			if (low[v] >= dfn[u] && u != r) cut[u] = 1,cnt ++;//不是根而且他的孩子无法跨越他回到祖先  割点
			if (r == u)child++; //如果是搜索树的根，统计孩子数目
		}
		low[u] = min(low[u], dfn[v]);//已经搜索过了
	}
	if (child >= 2 && u == r)cut[r] = 1,cnt ++;
}
void solve()
{
    cin >> n >> m;
    for(int i=1;i<=m;i++)
    {
        int u,v;
        cin >> u >> v;
        g[u].push_back(v),g[v].push_back(u);
    }
    pre(i,1,n)
    {
        if(!dfn[i])
            tarjan(i,i);
    }
    cout << cnt << endl;
    pre(i,1,n)
    {
        if(cut[i])  cout << i << " ";
    }
}

//#define LOCAL
signed main(){
    ios
    //fre
    #ifdef LOCAL
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
    auto start = std::chrono::high_resolution_clock::now();
    #endif
    
    int _=1;
    // cin>>_;
    while(_--)solve();
    
    #ifdef LOCAL
    auto end = std::chrono::high_resolution_clock::now();
    cout << "Execution time: "
         << std::chrono::duration_cast<std::chrono::milliseconds>(end - start).count()
         << " ms" << '\n';
    #endif
    return 0;
}