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 NMAX = 100005;
vector<int> graph[NMAX];
bool vis[NMAX];
int dist[NMAX];
int father[NMAX];
void dfs(int node) {
  vis[node] = true;
  for (auto it : graph[node])
    if (!vis[it]) {
      dist[it] = 1 + dist[node];
      father[it] = node;
      dfs(it);
    }
  int best_node;
  int best_dist = NMAX;
  for (auto it : graph[node])
    if (dist[it] < best_dist) {
      best_dist = dist[it];
      best_node = it;
    }
  cout << dist[node] - best_dist + 1 << '\n';
  while (node != best_node) {
    cout << node << ' ';
    node = father[node];
  }
  cout << node << endl;
  exit(0);
}
int main() {
  ios_base ::sync_with_stdio(false);
  int n, m, k;
  cin >> n >> m >> k;
  int a, b;
  while (m--) {
    cin >> a >> b;
    graph[a].push_back(b);
    graph[b].push_back(a);
  }
  dfs(1);
  return 0;
}
