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;
using ll = long long;
const int sz = 1e5 + 10;
const int szz = 1e6 + 5e5 + 10;
const ll mod = 1e9 + 7;
int n, m, k;
stack<int> st;
vector<int> G[sz], ans;
int vis[sz], cyc[sz], found;
void dfs(int u = 1, int dep = 1) {
  if (found) return;
  vis[u] = dep;
  cyc[u] = 1;
  st.push(u);
  for (const int& v : G[u]) {
    if (!vis[v]) {
      dfs(v, dep + 1);
    } else if (cyc[v]) {
      if (vis[u] - vis[v] >= k && !found) {
        int x;
        do {
          if (st.empty()) break;
          x = st.top();
          st.pop();
          ans.push_back(x);
        } while (x != v);
        found = 1;
      }
    }
  }
  cyc[u] = 0;
  if (!st.empty()) st.pop();
}
int main() {
  while (cin >> n >> m >> k) {
    memset(vis, 0, sizeof vis);
    memset(cyc, 0, sizeof cyc);
    found = 0;
    for (int i = 0; i <= n; i++) G[i].clear();
    for (int i = 0; i < m; i++) {
      int x, y;
      cin >> x >> y;
      G[x].push_back(y);
      G[y].push_back(x);
    }
    ans.clear();
    dfs();
    cout << ans.size() << "\n";
    for (const int& x : ans) cout << x << " ";
    cout << "\n";
  }
  return 0;
}
