// Success consists of going from failure to failure without loss of enthusiasm
#include <bits/stdc++.h>
using namespace std;
#define nl '\n'
#define pb push_back
#define mp make_pair
#define f first
#define s second
#define sz(x) int(x.size())
template<class T> using V = vector<T>;
using vi = V<int>;
using ll = long long;
using pi = pair<int, int>;
using vpi = V<pi>;
using vl = V<ll>;
int main() {
cin.tie(0)->sync_with_stdio(0);
int N, M; cin >> N >> M;
if (M != N - 1) {
cout << "NO" << nl;
return 0;
}
V<vi> adj(N); for(int i = 0; i < M; i++) {
int u, v; cin >> u >> v; --u, --v;
adj[u].pb(v);
adj[v].pb(u);
}
vi leaf(N);
int R = 0;
function<void(int, int)> dfs = [&](int u, int p) {
leaf[u] = 1;
int ADJ = 0;
for(auto& v : adj[u]) if (v != p) {
leaf[u] = 0; dfs(v, u);
if (!leaf[v]) ADJ++;
}
if (!leaf[u] && ADJ == 1) R = u;
if (ADJ > 2) {
cout << "NO" << nl;
exit(0-0);
}
};
dfs(0, -1);
vi ans;
function<void(int, int)> answer = [&](int u, int p) {
bool has = 0;
for(auto& v : adj[u]) if (v != p) {
if (leaf[v]) { has = 1; continue; }
else {
dfs(v, u);
}
}
ans.pb(u);
if (has) ans.pb(u);
};
answer(R, -1);
cout << "YES" << nl;
cout << sz(ans) << nl;
for(auto& x : ans) cout << x + 1 << " ";
cout << nl;
exit(0-0);
}
