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 long long MX = 1e18;
const long long MD = 1e9 + 7;
const int N = 1e5 + 5;
int n, m, k, vis[100005];
vector<int> gr[100005], ans;
void dfs(int x, int p) {
  vis[x] = 1;
  ans.push_back(x);
  for (int i = 0; i < gr[x].size(); i++) {
    int u = gr[x][i];
    if (vis[u] == 1 && ans.size() >= k + 1 && u != p) {
      vector<int> a;
      for (int j = ans.size() - 1; j >= 0; j--) {
        a.push_back(ans[j]);
        if (ans[j] == u) break;
      }
      if (a.size() >= k + 1) {
        cout << a.size() << endl;
        for (int j = 0; j < a.size(); j++) cout << a[j] << ' ';
        exit(0);
      }
    }
    if (vis[u] == 0) dfs(u, x);
  }
  ans.pop_back();
  vis[x] = 2;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    gr[x].push_back(y);
    gr[y].push_back(x);
  }
  for (int i = 1; i <= n; i++) {
    if (vis[i] == 0) dfs(i, 0);
  }
}
