/*
CEOI 2016 Potemkin Cycle
- Let each edge (u, v) represent a node (uv) in graph H
- There is an edge between (uv) and (vw) in H iff (uw) doesn't exist
- This means that there are no cycles of length 3 in H
- We can construct H in O(NR) time
- An answer exists iff a cycle exists in H; run DFS to find a cycle
*/
#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
int u[200001], v[200001], adj[1001][1001];
vector<int> graph[100001], stck;
int visited[100001];
void dfs(int node, int root, int parent = -1) {
if (visited[node] == root) {
while (stck.back() != node) {
cout << u[stck.back()] << ' ';
stck.pop_back();
}
cout << u[node];
exit(0);
}
if (visited[node]) return;
stck.push_back(node);
visited[node] = root;
for (int i : graph[node]) if (i != parent) dfs(i, root, node);
}
int main() {
cin.tie(0)->sync_with_stdio(0);
int n, r;
cin >> n >> r;
for (int i = 1; i <= r; i++) {
cin >> u[i] >> v[i];
u[i + r] = v[i], v[i + r] = u[i];
adj[u[i]][v[i]] = i, adj[v[i]][u[i]] = i + r;
}
for (int i = 1; i <= 2 * r; i++) {
for (int j = 1; j <= n; j++) if (j != u[i]) {
if (adj[v[i]][j] && !adj[u[i]][j])
graph[i].push_back(adj[v[i]][j]);
}
}
for (int i = 1; i <= 2 * r; i++) if (!visited[i]) dfs(i, i);
cout << "no";
return 0;
}
