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 inf = (int)1e9;
const int mod = inf + 7;
const double eps = 1e-9;
const double pi = acos(-1.0);
int n, m, k;
vector<int> g[100100], t;
int used[100100];
set<int> second[100100];
void dfs(int v) {
  used[v] = 1;
  t.push_back(v);
  bool was = 0;
  for (int i = 0; i < g[v].size(); ++i) {
    int to = g[v][i];
    if (!used[to]) {
      was = 1;
      dfs(to);
    }
  }
  if (!was) {
    int pos = 0;
    for (int i = 0; i < t.size(); ++i) {
      if (second[t[i]].find(v) != second[t[i]].end()) {
        pos = i;
        break;
      }
    }
    cout << t.size() - pos << endl;
    for (int i = pos; i < t.size(); ++i) {
      cout << t[i] << " ";
    }
    exit(0);
  }
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  cin >> n >> m >> k;
  for (int i = 0, u, v; i < m; ++i) {
    cin >> u >> v;
    second[u].insert(v);
    second[v].insert(u);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1);
}
