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;
void FastIO() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
}
vector<int> path, marked;
vector<vector<int> > V;
void dfs(int u) {
  marked[u] = 1;
  path.push_back(u);
  for (int i = 0; i < V[u].size(); i++) {
    int v = V[u][i];
    if (!marked[v]) {
      dfs(v);
      break;
    }
  }
}
int main() {
  FastIO();
  int n, m, k, a, b, r = -1;
  cin >> n >> m >> k;
  marked.assign(n + 1, 0);
  V.assign(n + 1, vector<int>());
  for (int i = 0; i < (int)m; i++) {
    cin >> a >> b;
    V[a].push_back(b);
    V[b].push_back(a);
  }
  dfs(1);
  int u = path.back();
  for (int i = path.size() - k - 1; i >= 0; i--) {
    cerr << "path[i]"
         << ": " << path[i] << '\n';
    ;
    for (int j = 0; j < V[u].size(); j++) {
      int v = V[u][j];
      if (v == path[i]) {
        cerr << "u"
             << ": " << u << " | "
             << "v"
             << ": " << v << '\n';
        ;
        r = path.size() - i;
        break;
      }
    }
    if (r > 0) break;
  }
  cout << r << '\n';
  for (int i = 0; i < r; i++) {
    cout << path.back() << " ";
    path.pop_back();
  }
  cout << '\n';
  return 0;
}
