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 MAXN = 1000 * 100 + 10;
int h[MAXN], par[MAXN];
bool mark[MAXN];
vector<int> adj[MAXN];
int uu = -1, ww = -1;
int n, m, t;
void dfs(int v) {
  mark[v] = true;
  int m = INT_MAX;
  int x = 0;
  for (int i = 0; i < adj[v].size(); i++) {
    int y = adj[v][i];
    if (!mark[y]) {
      x++;
      par[y] = v;
      h[y] = h[v] + 1;
      dfs(y);
      if (uu != -1 && ww != -1) return;
    }
    if (h[y] < m) {
      m = h[y];
      t = y;
    }
  }
  if (x == 0) {
    uu = v;
    ww = t;
  }
  return;
}
void pp(int x) {
  if (x == ww) {
    cout << ww + 1 << endl;
    return;
  }
  cout << x + 1 << " ";
  pp(par[x]);
  return;
}
int main() {
  int k;
  cin >> n >> m >> k;
  int u, v;
  for (int i = 0; i < m; i++) {
    cin >> u >> v;
    u--;
    v--;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  par[0] = -1;
  dfs(0);
  cout << h[uu] - h[ww] + 1 << endl;
  pp(uu);
  return 0;
}
