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<int> adj[100044];
bool vis[100001];
int dis[100001];
stack<int> nodes;
bool finish = false;
void print_solve(int stop, int res);
int dis_counter = 0;
void dfs(int v, int parent) {
  if (finish) return;
  vis[v] = true;
  dis[v] = dis_counter++;
  nodes.push(v);
  for (int i = 0; i < adj[v].size(); i++) {
    if (!vis[adj[v][i]]) {
      dfs(adj[v][i], v);
    } else if (adj[v][i] != parent && !finish &&
               (dis[v] - dis[adj[v][i]]) >= k) {
      print_solve(adj[v][i], dis[v] - dis[adj[v][i]]);
      finish = true;
      return;
    }
  }
  if (!finish) nodes.pop();
}
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  for (int i = 1; i <= n; i++) {
    if (finish) break;
    if (!vis[i]) dfs(i, i);
  }
  return 0;
}
void print_solve(int stop, int res) {
  cout << res + 1 << endl;
  while (!nodes.empty()) {
    cout << nodes.top() << " ";
    if (nodes.top() == stop) break;
    nodes.pop();
  }
}
