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;
int n, e, k;
vector<int> g[100010];
map<int, int> m;
bool usato[100010];
vector<int> path;
void dfs(int v) {
  usato[v] = true;
  path.push_back(v);
  m[v] = path.size();
  for (int i = 0; i < g[v].size(); i++)
    if (!usato[g[v][i]]) {
      dfs(g[v][i]);
      return;
    }
}
int main() {
  cin >> n >> e >> k;
  for (int i = 0; i < e; i++) {
    int a, b;
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(1);
  int v = path[path.size() - 1];
  int p = g[v][0];
  for (int i = 1; i < g[v].size(); i++)
    if (m[p] > m[g[v][i]]) p = g[v][i];
  vector<int> sol;
  int index = 0;
  while (path[index] != p) index++;
  for (index; index < path.size(); index++) sol.push_back(path[index]);
  cout << sol.size() << endl;
  for (int i = 0; i < sol.size(); i++) cout << sol[i] << " ";
  cout << endl;
  return 0;
}
