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> adj[100005], v;
;
bool vis[100005];
int par[100005] = {}, k, c[100005] = {}, st[100005] = {}, l = 0, x;
int dfs(int s) {
  vis[s] = true;
  st[s] = 1;
  if (l != 0) return x;
  for (int i = 0; i < adj[s].size(); i++) {
    if (vis[adj[s][i]] == false) {
      c[adj[s][i]] = c[s] + 1;
      par[adj[s][i]] = s;
      dfs(adj[s][i]);
    } else if (st[adj[s][i]] == 1 && c[s] - c[adj[s][i]] > 1) {
      if (c[s] - c[adj[s][i]] + 1 >= k + 1) {
        l = c[s] - c[adj[s][i]] + 1;
        x = s;
        return x;
      }
    }
  }
  if (l != 0) return x;
  cout << "aya";
  st[s] = -1;
  return 0;
}
int main() {
  int m, n, i, a, b, d;
  cin >> n >> m >> k;
  for (i = 0; i < m; i++) {
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  for (i = 1; i <= n; i++) {
    if (vis[i] == false) d = dfs(i);
    if (d != 0) {
      cout << l << endl;
      while (--l) {
        cout << d << " ";
        d = par[d];
      }
      cout << d;
      break;
    }
  }
}
