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;
long long v, e, k, from, to, start = 0, End = 0, res = 0;
map<long long, vector<long long>> adj;
vector<long long> vis(200010), len(200010);
map<long long, long long> parent;
void dfs(long long node, long long level) {
  len[node] = level;
  vis[node] = 1;
  for (int i = 0; i < adj[node].size(); i++) {
    long long child = adj[node][i];
    if (vis[child] && level - len[child] >= k) {
      start = child;
      End = node;
      parent[child] = node;
    }
    if (!vis[child] && start + End == 0) {
      dfs(child, level + 1);
      parent[child] = node;
    }
  }
}
void print_path(long long start, long long End) {
  cout << End << " ";
  if (End == start) return;
  print_path(start, parent[End]);
}
int main() {
  cin >> v >> e >> k;
  for (int i = 1; i <= e; i++) {
    cin >> from >> to;
    adj[from].push_back(to);
    adj[to].push_back(from);
  }
  dfs(1, 0);
  cout << len[End] - len[start] + 1 << endl;
  print_path(start, End);
  return 0;
}
