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, m, k;
vector<vector<int> > g;
vector<int> ans;
bool used[100500];
bool flag;
int fl;
int counter;
void dfs(int v) {
  used[v] = true;
  int c = 0;
  for (int to : g[v]) {
    if (flag) break;
    if (!used[to]) {
      dfs(to);
      c++;
    }
  }
  if (!flag && c == 0) {
    ans.push_back(v);
    fl = v;
    flag = true;
    return;
  }
  if (counter > 0) {
    for (int to : g[fl]) {
      if (v == to) {
        counter--;
        ans.push_back(v);
        return;
      }
    }
  }
  if (counter > 0) {
    ans.push_back(v);
  }
}
int main() {
  cin >> n >> m >> k;
  g.resize(n + 1);
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  memset((used), 0, sizeof(used));
  flag = false;
  fl = -1;
  counter = k;
  dfs(1);
  cout << ans.size() << endl;
  for (int i : ans) cout << i << " ";
  return 0;
}
