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, ans;
vector<int> graph[100001];
int vis[100001], d[100001];
bool flag = false;
vector<int> res;
void dfs(int v, int par) {
  vis[v] = 1;
  for (int i : graph[v]) {
    if (i == par) continue;
    if (vis[i] == 1) {
      if (d[v] - d[i] + 1 > k) {
        flag = true;
        ans = d[v] - d[i];
        res.push_back(v);
        return;
      }
      continue;
    }
    d[i] = 1 + d[v];
    dfs(i, v);
    if (flag) {
      if (ans) {
        res.push_back(v);
        ans -= 1;
      }
      return;
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  memset(vis, 0, 100001);
  memset(d, 0, 100001);
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    graph[a].push_back(b);
    graph[b].push_back(a);
  }
  dfs(1, 0);
  cout << res.size() << '\n';
  for (int i = 0; i < res.size(); i++) cout << res[i] << ' ';
  return 0;
}
