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;
bool vis[100001];
int pi[100001];
int level[100001];
list<int> adj[100001];
void dfs(int s) {
  vis[s] = 1;
  bool flag = 1;
  for (list<int>::iterator it = adj[s].begin(); it != adj[s].end(); ++it)
    if (!vis[*it]) {
      level[*it] = level[s] + 1;
      pi[*it] = s;
      dfs(*it);
      flag = 0;
    }
  if (flag) {
    int least = s;
    stack<int> stk;
    for (list<int>::iterator it = adj[s].begin(); it != adj[s].end(); ++it)
      if (level[least] > level[(*it)]) least = *it;
    while (s != least) {
      stk.push(s);
      s = pi[s];
    }
    stk.push(s);
    cout << stk.size() << "\n";
    while (!stk.empty()) {
      cout << stk.top() << " ";
      stk.pop();
    }
    exit(0);
  }
}
int main() {
  int n, m, k, i, l, r;
  cin >> n >> m >> k;
  for (i = 1; i <= m; ++i) {
    cin >> l >> r;
    adj[l].push_back(r);
    adj[r].push_back(l);
  }
  for (i = 1; i <= n; ++i)
    if (!vis[i]) {
      pi[i] = -1;
      dfs(i);
    }
  return 0;
}
