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 inf = 1e18 + 10;
const long long mod = 1e9 + 7;
const long long maxn = 1e5 + 10, sq = 430 + 10, maxt = 1e5 + 10,
                mlg = (1 << 18) + 10;
vector<long long> nei[maxn], ans;
long long n, m, k, par[maxn], h[maxn], mark[maxn];
void come_back(long long u, long long v) {
  ans.push_back(u);
  if (u != v) {
    come_back(par[u], v);
  }
}
bool dfs(long long u) {
  mark[u] = 1;
  for (auto v : nei[u]) {
    if (!mark[v]) {
      par[v] = u;
      h[v] = h[u] + 1;
      if (dfs(v)) {
        return 1;
      }
    } else {
      if (h[u] - h[v] >= k) {
        come_back(u, v);
        return 1;
      }
    }
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> m >> k;
  for (int i = 1; i <= m; i++) {
    long long a, b;
    cin >> a >> b;
    nei[a].push_back(b);
    nei[b].push_back(a);
  }
  dfs(1);
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) {
    cout << ans[i] << ' ';
  }
  cout << endl;
}
