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<long long> v[123456];
bool vis[123456];
long long h[123456];
long long par[123456];
long long n, m, k;
bool cyc = false;
long long idx = -1;
long long ans;
void dfs(int node) {
  if (cyc) return;
  vis[node] = true;
  int z = v[node].size();
  int nd;
  for (int i = 0; i < z; i++) {
    nd = v[node][i];
    if (vis[nd]) {
      long long t = h[node] - h[nd] + 1;
      if (t >= k + 1) {
        cyc = true;
        idx = node;
        ans = t;
        return;
      }
      continue;
    }
    par[nd] = node;
    h[nd] = h[node] + 1;
    dfs(nd);
  }
  return;
}
int main() {
  cin >> n >> m >> k;
  int x, y;
  for (int i = 0; i < m; i++) {
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  h[1] = 1;
  dfs(1);
  cout << ans << endl;
  for (int i = 0; i < ans; i++) {
    cout << idx << " ";
    idx = par[idx];
  }
  cout << endl;
  return 0;
}
