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 N = 1E6 + 7;
const int INF = 1E9;
vector<int> res;
vector<int> adj[N];
int parent[N];
int n, m, k;
int f[N];
void DFS(int u) {
  for (auto v : adj[u])
    if (parent[v] == 0) {
      f[v] = f[u] + 1;
      parent[v] = u;
      DFS(v);
    }
}
int main() {
  ios::sync_with_stdio(NULL);
  cin.tie(NULL);
  cin >> n >> m >> k;
  for (int i = 1; i <= m; ++i) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  parent[1] = -1;
  DFS(1);
  for (int u = 2; u <= n; ++u) {
    for (auto v : adj[u])
      if (parent[u] != v && f[v] < f[u]) {
        int len = f[u] - f[v] + 1;
        if (len >= k + 1) {
          int v_2 = u;
          cout << len << "\n";
          while (v_2 != v) {
            cout << v_2 << " ";
            v_2 = parent[v_2];
          }
          return cout << v, 0;
        }
      }
  }
}
