// 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 sz(x) int(x.size())
template<class T> using V = vector<T>;
using vi = V<int>;
int main() {
cin.tie(0)->sync_with_stdio(0);
int N, M; cin >> N >> M;
V<vi> adj(N); vi in(N), path(N);
for(int i = 0; i < N - 1; i++) {
int u, v; cin >> u >> v; --u, --v;
adj[u].pb(v);
adj[v].pb(u);
in[u]++, in[v]++;
}
if (N == 1) {
cout << "YES" << nl;
cout << 1 << nl;
cout << 1 << nl;
exit(0-0);
}
if (N == 2) {
cout << "YES" << nl;
cout << 2 << nl;
cout << 1 << " " << 1 << nl;
exit(0-0);
}
if (M != N - 1) {
cout << "NO" << nl;
exit(0-0);
}
for(int u = 0; u < N; u++) if (in[u] == 1) {
for(auto& v : adj[u]) path[v] = 1;
}
for(int u = 0; u < N; u++) if (!path[u]) {
for(auto& v : adj[u]) in[v]--;
}
vi ans;
function<void(int)> dfs = [&](int u) {
ans.pb(u); path[u] = 0;
int chd = 0;
for(auto& v : adj[u]) if (path[v]) {
dfs(v); chd++;
}
if (chd >= 2) {
cout << "NO" << nl;
exit(0-0);
}
ans.pb(u);
};
for(int u = 0; u < N; u++) if (path[u] && in[u] <= 1) dfs(u);
if (sz(ans) == 0) {
ans.pb(0); ans.pb(0);
}
cout << "YES" << nl;
cout << sz(ans) << nl;
for(auto& x : ans) cout << x + 1 << " ";
cout << nl;
exit(0-0);
}
