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;
vector<int> gr[100005];
bool visit[100005], f = 0;
int len, par[100005], st[100005], K;
void dfs(int u) {
  int i, v;
  visit[u] = 1;
  for (i = 0; i < gr[u].size(); i++) {
    if (f) break;
    if (!visit[gr[u][i]])
      par[gr[u][i]] = u, dfs(gr[u][i]);
    else if (par[u] != gr[u][i]) {
      len = 0;
      v = u;
      st[len++] = gr[u][i];
      while (v != gr[u][i]) {
        st[len++] = v;
        v = par[v];
      }
      if (len > K) f = 1;
    }
  }
}
int main() {
  int n, i, j, a[100005], m, u, v;
  cin >> n >> m >> K;
  while (m--) {
    cin >> u >> v;
    gr[u].push_back(v);
    gr[v].push_back(u);
  }
  memset(visit, 0, sizeof(visit));
  for (i = 1; i <= n; i++) {
    if (visit[i]) continue;
    if (f) break;
    dfs(i);
  }
  cout << len << endl;
  for (i = 0; i < len; i++) cout << st[i] << " ";
  return 0;
}
