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 MAX = 100005;
const long long MOD = 1000000007;
const double PI = 2 * acos(0.0);
const double EPS = 0.00000000001;
vector<int> adj[MAX], order;
int n, m, k, x, y, f = -1, l = -1, z = -1;
unordered_map<int, int> dist, visit;
bool flag = false;
void dfs(int u, int c) {
  visit[u] = 1;
  dist[u] = c;
  order.push_back(u);
  for (int i = 0; i < adj[u].size(); i++) {
    int v = adj[u][i];
    if (flag)
      break;
    else if (visit[v] == 0)
      dfs(v, c + 1);
    else {
      if (abs(dist[u] - dist[v]) + 1 >= k + 1) {
        f = v, l = u, z = abs(dist[u] - dist[v]) + 1;
        flag = true;
        break;
      }
    }
  }
  if (!flag) order.pop_back();
}
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  dist[1] = 1;
  dfs(1, 1);
  cout << z << endl;
  for (int i = 0; i < order.size(); i++) {
    if (order[i] != f)
      continue;
    else {
      while (i < order.size()) {
        cout << order[i] << " ";
        if (order[i] == l) break;
        i++;
      }
      break;
    }
  }
  return 0;
}
