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 MAX = 2e5 + 35;
const long long mod = 1e9 + 7;
const double eps = 1e-6;
using namespace std;
vector<int> adj[MAX];
int vis[MAX];
map<pair<int, int>, bool> mp;
int maap[MAX];
int c = 0;
int cc;
void dfs(int i) {
  maap[c] = i;
  c++;
  vis[i] = 1;
  bool ok = 1;
  for (int j = 0; j < adj[i].size(); j++) {
    int u = adj[i][j];
    if (!vis[u]) {
      ok = 0;
      dfs(u);
    }
  }
  if (ok && !cc) cc = c;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie();
  cout.tie();
  int n, m, k;
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
    mp[{x, y}] = mp[{y, x}] = 1;
  }
  dfs(1);
  int x = maap[cc - 1];
  int in = -1;
  for (int i = 0; i < cc; i++) {
    int y = maap[i];
    if (mp[{x, y}]) {
      in = i;
      break;
    }
  }
  cout << cc - in << "\n";
  for (int i = in; i < cc; i++) {
    cout << maap[i] << " ";
  }
  return 0;
}
