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;
vector<int> G[100010];
int n, m, k;
bool dfs(int v, vector<int>& res, vector<int>& visited) {
  for (auto u : G[v]) {
    if (visited[u] and res.size() - visited[u] + 1 > k) {
      cout << res.size() - visited[u] + 1 << endl;
      for (int i = visited[u] - 1; i < res.size(); ++i)
        cout << res[i] + 1 << ' ';
      cout << endl;
      return true;
    } else if (!visited[u]) {
      res.push_back(u);
      visited[u] = res.size();
      if (dfs(u, res, visited)) return true;
      res.pop_back();
      visited[u] = 0;
    }
  }
  return false;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> k;
  while (m--) {
    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) {
    vector<int> res, visited(n);
    res.push_back(i);
    visited[i] = 1;
    if (dfs(i, res, visited)) break;
  }
}
