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 used[100001];
vector<int> adj[100001];
bool solve(vector<int>& ans, int n) {
  if (ans.size() >= n) {
    vector<int> v1 = adj[ans[0]];
    bool good = false;
    for (int i = 0; i < v1.size(); i++) {
      if (ans.back() == v1[i]) good = true;
    }
    if (good) return true;
  }
  vector<int> neighbors = adj[ans.back()];
  for (int i = 0; i < neighbors.size(); i++) {
    if (used[neighbors[i]] == 0) {
      used[neighbors[i]] = 1;
      ans.push_back(neighbors[i]);
      if (solve(ans, n)) return true;
      ans.pop_back();
      used[neighbors[i]] = 0;
    }
  }
  return false;
}
int main() {
  long n, m, k;
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  vector<int> ans;
  ans.push_back(1);
  used[1] = 1;
  solve(ans, k + 1);
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i] << " ";
  }
  cout << endl;
}
