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 dis[100005];
vector<int> adj[100005], path;
void DFS(int u) {
  path.push_back(u);
  for (int i = 0; i < (adj[u].size()); i++) {
    int v = adj[u][i];
    if (dis[v] == -1) {
      dis[v] = dis[u] + 1;
      DFS(v);
      break;
    }
  }
  return;
}
int main() {
  int n, m, k, x, y;
  memset(dis, -1, sizeof(dis));
  scanf("%d %d %d", &n, &m, &k);
  while (m--) {
    scanf("%d %d", &x, &y);
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  dis[1] = 0;
  DFS(1);
  int e = path.back(), s = adj[e][0];
  for (int i = 0; i < adj[e].size(); i++)
    if (dis[s] > dis[adj[e][i]]) s = adj[e][i];
  int id;
  for (int i = 0; i < path.size(); i++)
    if (path[i] == s) {
      id = i;
      break;
    }
  cout << path.size() - id << endl;
  printf("%d", s);
  for (int i = id + 1; i < path.size(); i++) printf(" %d", path[i]);
  printf("\n");
  return 0;
}
