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 maxn = 111111;
int n, m, k, path[maxn], belong[maxn], pathLen;
vector<int> g[maxn];
bool vis[maxn];
void dfs(int pos) {
  vis[pos] = true;
  path[++pathLen] = pos;
  belong[pos] = pathLen;
  for (vector<int>::iterator it = g[pos].begin(); it != g[pos].end(); it++) {
    int v = (*it);
    if (!vis[v]) {
      dfs(v);
      break;
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  memset(vis, false, sizeof(vis));
  cin >> n >> m >> k;
  for (int i = 1; i <= m; ++i) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1);
  int pos, minNum = 0x7FFFFFFF / 2;
  for (vector<int>::iterator it = g[path[pathLen]].begin();
       it != g[path[pathLen]].end(); it++) {
    if (belong[(*it)] < minNum && vis[(*it)]) {
      minNum = belong[(*it)];
      pos = (*it);
    }
  }
  cout << pathLen - minNum + 1 << endl;
  for (int i = minNum; i <= pathLen - 1; ++i) cout << path[i] << " ";
  cout << path[pathLen] << endl;
  return 0;
}
