## Algorithm

D. Cycle in Graph
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

You've got a undirected graph G, consisting of n nodes. We will consider the nodes of the graph indexed by integers from 1 to n. We know that each node of graph G is connected by edges with at least k other nodes of this graph. Your task is to find in the given graph a simple cycle of length of at least k + 1.

simple cycle of length d (d > 1) in graph G is a sequence of distinct graph nodes v1, v2, ..., vd such, that nodes v1 and vd are connected by an edge of the graph, also for any integer i (1 ≤ i < d) nodes vi and vi + 1 are connected by an edge of the graph.

Input

The first line contains three integers nmk (3 ≤ n, m ≤ 105; 2 ≤ k ≤ n - 1) — the number of the nodes of the graph, the number of the graph's edges and the lower limit on the degree of the graph node. Next m lines contain pairs of integers. The i-th line contains integers aibi (1 ≤ ai, bi ≤ nai ≠ bi) — the indexes of the graph nodes that are connected by the i-th edge.

It is guaranteed that the given graph doesn't contain any multiple edges or self-loops. It is guaranteed that each node of the graph is connected by the edges with at least k other nodes of the graph.

Output

In the first line print integer r (r ≥ k + 1) — the length of the found cycle. In the next line print r distinct integers v1, v2, ..., vr (1 ≤ vi ≤ n) — the found simple cycle.

It is guaranteed that the answer exists. If there are multiple correct answers, you are allowed to print any of them.

Examples
input
Copy
`3 3 21 22 33 1`
output
Copy
`31 2 3 `
input
Copy
`4 6 34 31 21 31 42 32 4`
output
Copy
`43 4 1 2 `

## Code Examples

### #1 Code Example with C++ Programming

```Code - C++ Programming```

``````#include <bits/stdc++.h>

using namespace std;

int const N = 2e5 + 10;
int n, m, k;
bool vis[N];
vector<vector<int> > g;
vector<int> sol;

void DFS(int v, int src) {
vis[v] = true;
sol.push_back(v);

for(int i = 0; i < (int)g[v].size(); ++i) {
int u = g[v][i];
if(u == src && sol.size() >= k + 1) {
printf("%d\n", (int)sol.size());
for(int j = 0; j < (int)sol.size(); ++j)
printf("%d ", sol[j]);
puts("");
exit(0);
}
if(!vis[u])
DFS(u, src);
}

sol.pop_back();
}

int main() {
scanf("%d %d %d", &n, &m, &k);
g.resize(n + 10);
for(int i = 0, a, b; i < m; ++i) {
scanf("%d %d", &a, &b);
g[a].push_back(b);
swap(a, b);
g[a].push_back(b);
}
for(int i = 0; i < (int)g.size(); ++i)
sort(g[i].begin(), g[i].end());

for(int i = 1; i < n; ++i) {
memset(vis, false, sizeof vis);
DFS(i, i);
}

return 0;
}``````
Copy The Code &

Input

cmd
3 3 2
1 2
2 3
3 1

Output

cmd
3
1 2 3