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.

A 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 n, m, k (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 ai, bi (1 ≤ ai, bi ≤ n; ai ≠ 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

3 3 2
1 2
2 3
3 1


Output

3
1 2 3 

Input

4 6 3
4 3
1 2
1 3
1 4
2 3
2 4


Output

4
3 4 1 2 
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;
int n, m, k, dis[N], max_dis, dad[N], length = 1, cycle;
vector<int> adj[N], ans;
bool seen[N], way[N];
void DFS(int v) {
  seen[v] = true;
  for (int i = 0; i < adj[v].size(); i++)
    if (!seen[adj[v][i]]) {
      dis[adj[v][i]] = dis[v] + 1;
      max_dis = (dis[max_dis] > dis[v] + 1 ? max_dis : adj[v][i]);
      dad[adj[v][i]] = v;
      DFS(adj[v][i]);
    }
  return;
}
int main() {
  cin >> n >> m >> k;
  k++;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  DFS(0);
  ans.push_back(max_dis);
  for (int i = 0; i < adj[max_dis].size(); i++) way[adj[max_dis][i]] = true;
  cycle = max_dis;
  while (true) {
    ans.push_back(dad[cycle]);
    length++;
    cycle = dad[cycle];
    if (length >= k && way[cycle]) break;
  }
  cout << length << endl;
  for (int i = 0; i < ans.size(); i++) cout << ans[i] + 1 << " ";
  return 0;
}
