/*
尽可能减小问题规摸
分析必要条件，能不能得到充要条件
讨论最大/最小的元素的选择
想想简化的问题怎么做。或者是 x=0/x=1 这样的特殊情况。
拆贡献
容斥
分析答案上下界 
化简状态
正难则反
分析单调性，凸性
可以尝试网络流、高斯消元

专注，冷静。
*/
#include <bits/stdc++.h>
#define rep(i, f, t) for(int i(f); i <= t; ++i)
#define re(i, t) for(int i(1); i <= t; ++i)
#define per(i, t, f) for(int i(t); i >= f; --i)
#define pe(i, t) for(int i(t); i >= 1; --i) 
#define nx(i, u) for(int i(head[u]); i; i = e[i].nxt)
typedef long long ll;
typedef unsigned long long ull; 
using namespace std;
typedef pair <int, int> pii;
#define pb push_back
#define fi first
#define se second
#define ix(l, r) ((l + r) | (l != r))  
#define mp(i, j) (make_pair(i, j))
// #define int long long
#define inf (int)(1e9+7)
#define INF 0x3f3f3f3f3f3f3f3f
namespace IO {
char buf[1 << 21], *p1 = buf, *p2 = buf, buf1[1 << 21];
inline char gc() {return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++;}
template<class I>
inline void read(I &x) {x = 0;I f = 1;char c = gc();while (c < '0' || c > '9') {if (c == '-') {f = -1;} c = gc();}while (c >= '0' && c <= '9') {x = x * 10 + c - '0';c = gc();}x *= f;}
template<class I>
inline void write(I x) {if (x == 0) {putchar('0');return;}I tmp = x > 0 ? x : -x;if (x < 0) {putchar('-');}int cnt = 0;while (tmp > 0) {buf1[cnt++] = tmp % 10 + '0';tmp /= 10;}while (cnt > 0)putchar(buf1[--cnt]);}
#define outn(x) write(x), putchar('\n')
#define out(x) write(x), putchar(' ')
} using namespace IO;
#define look_memory cerr<<abs(&sT-&eD)/1024.0/1024<<'\n'
int sT;     


const int N = 4e5 + 5;
int n, m, u, v;

struct node {
	int to, nxt;
}e[N*5];

int head[N], cnt;

void add (int u ,int v) {
	e[++cnt] = (node){v, head[u]};
	head[u] = cnt;
}

int fath[N];
int deg[N];

int find (int u) {
	return (fath[u] == u) ? (u) : (fath[u] = find(fath[u]));
}

bool check (int lim) {
	re(i, n) fath[i] = i;
	re(u, n) {
		nx(i, u) {
			int v = e[i].to;
			if(u == lim || v == lim) continue;
			if(u < v) continue;
			int x = find(u), y = find(v);
			if(x == y) return 0;
			fath[x] = y;
		}
	}
	return 1;
}

vector<int>t1, t2, ans;
int cyc, vis[N],dfn[N], tot;

void dfs (int u, int fa) {
	vis[u] = 1;
	dfn[u] = ++tot;
	nx(i, u) {
		int v = e[i].to;
		if(v == fa) continue;
		if(!vis[v]) {
			fath[v] = u;
			dfs(v, u);
		}
		else if(vis[v] == 1) {
			int now = u;
			++cyc;
			while(now != v) {
				++deg[now];
				now = fath[now];
			}
			++deg[v];
		}
		if(cyc >= 2) return ;
	}
	vis[u] = 2;
}

int main () {
	read(n),read(m);
	re(i, m) {
		read(u),read(v);
		add(u,v),add(v, u);
		++deg[u], ++deg[v];
	}
	re(i, n) {
		if(deg[i] & 1) {
			puts("0");
			return 0;
		}
		deg[i] = 0;
	}
	if(n == m) {
		outn(n);
		re(i, n) {
			out(i);
		}
		return 0;
	}	
	dfs(1, 0);
	re(i, n) {
		if(deg[i] == 2) t1.pb(i);
	}
	sort(t1.begin(), t1.end(), [&](int u, int v){return dfn[u] < dfn[v];});
	int len = t1.size();
	if(len == 1) t2.pb(t1[0]);
	else t2.pb(t1[0]), t2.pb(t1[len-1]);
	for(int i : t2) if(check(i)) ans.pb(i);
	outn(ans.size());
	sort(ans.begin(), ans.end());
	for(int i : ans) {
		out(i);
	}
	return 0;
}