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 + 9;
const long long int mod = 1e9 + 7;
map<long long int, long long int> mpp, mp;
vector<long long int> u, v[N];
multiset<long long int> st, stt;
long long int c[N], t[] = {0, 2, 4, 9, 10, 11};
long long int a[N], n, m, k, vis[N], parent[N], level[N], length;
long long int dfs(long long int x, long long int papa, long long int lvl) {
  vis[x] = 1;
  parent[x] = papa;
  level[x] = lvl;
  for (auto i : v[x]) {
    if (!vis[i]) {
      long long int l = dfs(i, x, level[x] + 1);
      if (l) return l;
    } else if (vis[i] && i != papa) {
      if (lvl - level[i] + 1 >= k + 1) {
        length = lvl - level[i] + 1;
        return x;
      }
    }
  }
  vis[x] = 0;
  return 0;
}
void solve() {
  cin >> n >> m >> k;
  for (int i = 1; i <= m; ++i) {
    long long int x, y;
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  long long int cnt = 0;
  level[1] = 1;
  long long int l = dfs(1, 1, 1);
  cout << length << '\n' << l << " ";
  while (--length) {
    cout << parent[l] << " ";
    l = parent[l];
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t = 1;
  while (t--) solve();
}
