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>
const int MAX = 1e4 + 5;
const int INF = 1e9;
using namespace std;
pair<long long int, pair<long long int, long long int>> vec[MAX];
vector<long long int> adj[200005], aa;
long long int vis[200005];
long long int n, me = 0, ma = 0, a, b, c = 0, d = 0, k = 0, dep = 0, m = 0,
                 nod = 0, t, x, y;
long long int dpa[200005];
long long int dp[200005];
long long int p[200003];
unordered_map<long long int, unordered_map<long long int, long long int>> ab;
void dfs(long long int u) {
  vis[u] = 1;
  for (auto &i : adj[u]) {
    if (!vis[i]) {
      dp[i] = dp[u] + 1;
      dpa[i] = u;
      dfs(i);
    } else if (dp[i] - dp[u] >= k) {
      cout << dp[i] - dp[u] + 1 << endl;
      long long int kk = i;
      for (long long int j = 0; j <= dp[i] - dp[u]; j++) {
        cout << kk << " ";
        kk = dpa[kk];
      }
      exit(0);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> m >> k;
  for (long long int i = 0; i < m; i++) {
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  for (long long int i = 1; i <= n; i++) {
    if (!vis[i]) {
      dfs(i);
    }
  }
  return 0;
}
