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 = 100005;
int dr8[] = {0, 0, 1, -1, 1, 1, -1, -1};
int dc8[] = {1, -1, 0, 0, -1, 1, -1, 1};
int dr4[] = {0, 1, -1, 0};
int dc4[] = {1, 0, 0, -1};
vector<int> gr[N];
int vis[N], has[N];
vector<int> ans;
int k;
int dn = 0;
void dfs(int u, int d) {
  vis[u] = 1;
  if (dn == 1) return;
  ans.push_back(u);
  if (d > k && (has[u] == 1)) {
    dn = 1;
    cout << ans.size() << "\n";
    for (int i = 0; i < ans.size(); i++) {
      cout << ans[i] << " ";
    }
    return;
  }
  for (int i = 0; i < gr[u].size(); i++) {
    int v = gr[u][i];
    if (!vis[v]) {
      dfs(v, d + 1);
    }
  }
  ans.pop_back();
  vis[u] = 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m, x, y;
  cin >> n >> m >> k;
  memset(vis, 0, sizeof vis);
  memset(has, 0, sizeof has);
  for (int i = 0; i < m; i++) {
    cin >> x >> y;
    if (x == 1 || y == 1) {
      int mx = max(x, y);
      has[mx] = 1;
    }
    gr[x].push_back(y);
    gr[y].push_back(x);
  }
  dfs(1, 1);
  if (dn == 0) {
    for (int i = 1; i <= n; i++) {
      if (gr[i].size() > 2) {
        ans.clear();
        memset(has, 0, sizeof has);
        for (int j = 0; j < gr[i].size(); j++) {
          has[gr[i][j]] = 1;
        }
        dfs(i, 1);
        return 0;
      }
    }
  }
}
