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 NMAX = 1e5 + 7;
int n, m, k;
vector<int> g[NMAX];
pair<int, int> res = {-1, -1};
int len;
int vis[NMAX], tin[NMAX], tid[NMAX], par[NMAX];
int ideeees = 0;
void Dfs(int cur) {
  if (res.second >= 0) return;
  vis[cur] = 1;
  tin[cur] = ++ideeees;
  tid[ideeees] = cur;
  for (int i : g[cur]) {
    if (!vis[i]) {
      Dfs(i);
      par[i] = cur;
    } else {
      if (tin[cur] - tin[i] + 1 > k) {
        res = {i, cur};
        len = tin[cur] - tin[i] + 1;
        return;
      }
    }
  }
}
int main() {
  cin >> n >> m >> k;
  while (m--) {
    int a, b;
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  Dfs(1);
  cout << len << endl;
  int x = res.second;
  int y = res.first;
  while (x != y) {
    cout << x << " ";
    x = par[x];
  }
  cout << y;
  return 0;
}
