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 long long N = 1e5 + 7;
long long n, m, k;
vector<long long> g[N];
bool used[N];
long long h[N];
long long par[N];
long long som = -1;
void dfs(long long u) {
  used[u] = 1;
  bool leaf = 1;
  for (long long v : g[u]) {
    if (!used[v]) {
      h[v] = h[u] + 1;
      par[v] = u;
      dfs(v);
      leaf = 0;
    }
  }
  if (leaf) som = u;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> k;
  for (long long i = 0; i < m; ++i) {
    long long u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  par[1] = 1;
  dfs(1);
  long long u = som;
  assert(u != -1);
  long long up = -1;
  for (long long v : g[u]) {
    if (h[v] <= h[u] - k) {
      up = v;
    }
  }
  assert(up != -1);
  cout << h[u] - h[up] + 1 << '\n';
  long long t = u;
  while (t != up) {
    cout << t << ' ';
    t = par[t];
  }
  cout << t << '\n';
}
