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;
long long int n, m, k;
vector<vector<long long int> > g(100005, vector<long long int>());
vector<long long int> visit, level, parents;
void printpath(long long int t, long long int tt) {
  cout << level[tt] - level[t] + 1 << "\n";
  while (tt != t) {
    cout << tt << " ";
    tt = parents[tt];
  }
  cout << tt;
  exit(0);
}
long long int DFS(long long int x, long long int l, long long int pare) {
  visit[x] = 1;
  level[x] = l;
  parents[x] = pare;
  for (int i = 0; i < g[x].size(); i++) {
    if (!visit[g[x][i]]) DFS(g[x][i], l + 1, x);
    if (abs(level[g[x][i]] - level[x]) >= k) printpath(g[x][i], x);
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> m >> k;
  while (m--) {
    long long int x, y;
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  visit.assign(100005, false);
  level.assign(100005, 0);
  parents.assign(100005, -1);
  DFS(1, 1, 1);
}
