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;
int II() {
  int q;
  scanf("%d", &q);
  return q;
}
int dx[] = {0, -1, 0, 1};
int dy[] = {-1, 0, 1, 0};
int dxx[] = {0, -1, -1, -1, 0, 1, 1, 1};
int dyy[] = {-1, -1, 0, 1, 1, 1, 0, -1};
int n, m, k;
vector<int> v[100005], ans;
int dist[100005];
int vis[100005];
int temp;
bool dfs(int u, int di) {
  dist[u] = di;
  vis[u] = 1;
  for (auto it : v[u]) {
    if (vis[it] == 1) {
      if (abs(dist[it] - dist[u]) >= k) {
        temp = abs(dist[it] - dist[u]) - 1;
        ans.push_back(it);
        return 1;
      }
    }
    if (vis[it] == 0) {
      if (dfs(it, di + 1)) {
        if (temp >= 0) {
          ans.push_back(it);
          temp--;
        }
        return 1;
      }
    }
  }
  vis[u] = 2;
  return 0;
}
void solve() {
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int u, vv;
    cin >> u >> vv;
    v[u].push_back(vv);
    v[vv].push_back(u);
  }
  for (int i = 1; i <= n; i++) {
    if (vis[i] == 0) {
      if (dfs(i, 1)) break;
    }
  }
  cout << ans.size() << endl;
  for (auto it : ans) cout << it << " ";
}
int main() {
  int tt = 1;
  int cas = 1;
  while (tt--) {
    solve();
  }
}
