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> color, d, pr, cycle;
void dfs(int v, int p = -1) {
  color[v] = 1;
  pr[v] = p;
  if (p == -1)
    d[v] = 0;
  else
    d[v] = d[p] + 1;
  for (auto to : g[v]) {
    if (to == p) continue;
    if (color[to] == 0)
      dfs(to, v);
    else if (color[to] == 1 && cycle.empty()) {
      if (d[v] - d[to] >= k) {
        int now = v;
        cycle.clear();
        while (now != to) {
          cycle.push_back(now);
          now = pr[now];
        }
        cycle.push_back(to);
      }
    }
  }
  color[v] = 2;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> k;
  g.resize(n);
  color.resize(n);
  d.resize(n);
  pr.resize(n, -1);
  for (int i = 0; i < m; ++i) {
    int a, b;
    cin >> a >> b;
    --a;
    --b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  for (int i = 0; i < n; ++i) {
    if (color[i] == 0) dfs(i);
  }
  cout << cycle.size() << "\n";
  for (auto i : cycle) cout << i + 1 << " ";
  cout << "\n";
  return 0;
}
