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 MOD = 1e9 + 7;
const int MOD2 = 1e9 + 9;
const int PR = 727;
const int INF = INT_MAX;
const long long LINF = LLONG_MAX;
const int N = 1e5 + 20;
vector<int> G[N], p;
int n, m, k, f[N], tmp;
bool visited[N];
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> m >> k;
  for (int i = 0, v, u; i < m; i++) {
    cin >> v >> u;
    G[v].push_back(u);
    G[u].push_back(v);
  }
  int v = 1;
  while (1) {
    visited[v] = 1;
    f[v] = tmp++;
    p.push_back(v);
    for (int u : G[v])
      if (!visited[u]) v = u;
    if (visited[v]) break;
  }
  int idx = INF;
  for (int v : G[p.back()]) idx = min(idx, f[v]);
  cout << p.size() - idx << endl;
  for (int i = idx; i < p.size(); i++) cout << p[i] << ' ';
  cout << endl;
}
