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> l[100007];
int parent[100007];
int color[100007];
int k;
int p;
int dfs(int u, int c) {
  int sz = l[u].size();
  {
    for (int i = 0; i < sz; i++) {
      if (color[u] - color[l[u][i]] >= k && color[l[u][i]] != 0) {
        parent[l[u][i]] = u;
        return u;
      } else if (color[l[u][i]] == 0) {
        color[l[u][i]] = c + 2;
        parent[l[u][i]] = u;
        return dfs(l[u][i], c + 1);
      }
    }
  }
}
int main() {
  int n, m;
  cin >> n >> m >> k;
  int a, b;
  for (int i = 0; i < m; i++) {
    cin >> a >> b;
    l[a].push_back(b);
    l[b].push_back(a);
  }
  color[1] = 1;
  int u = dfs(1, 0);
  int v = parent[u];
  int cycle = 1;
  while (v != u) {
    v = parent[v];
    cycle++;
  }
  cout << cycle << endl;
  cout << u << " ";
  v = parent[u];
  while (v != u) {
    cout << v << " ";
    v = parent[v];
  }
}
