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 inf = 1e9, n, k, flag = 0;
const long long N = 2e6 + 10, mod = 1e9 + 7;
bool vis[N];
int low[N], dfn[N], Siz[N], tim = 0, root;
vector<int> adj[N], ans;
void dfs(int u, int p) {
  dfn[u] = low[u] = ++tim;
  for (auto v : adj[u]) {
    if (v == p) continue;
    if (!dfn[v]) {
      dfs(v, u);
      low[u] = min(low[u], low[v]);
    } else
      low[u] = min(dfn[v], low[u]);
  }
  Siz[low[u]]++;
}
void dfs2(int u) {
  vis[u] = 1;
  ans.push_back(u);
  if (flag) return;
  for (auto v : adj[u]) {
    if (!vis[v] && low[u] == low[v]) {
      dfs2(v);
    } else if (vis[v] && low[u] == low[v] && v == root && ans.size() >= k)
      flag = 1;
    if (flag) break;
  }
}
int main() {
  int u, v, m;
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1, -1);
  k++;
  memset(vis, 0, sizeof vis);
  int j = 0;
  for (int i = 1; i <= n; i++)
    if (Siz[i] >= k) {
      j = i;
      break;
    }
  for (int i = 1; i <= n; i++) {
    if (low[i] == j && low[i] == dfn[i]) {
      j = i;
      break;
    }
  }
  root = j;
  dfs2(j);
  cout << ans.size() << "\n";
  for (auto v : ans) cout << v << " ";
}
