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 n, m, k;
vector<int> e[100005];
int d[100005] = {};
int minLoc = -1;
bool dfs(int v, int depth) {
  int cnt = 0;
  d[v] = depth;
  for (int i = 0; i < e[v].size(); i++) {
    if (d[e[v][i]] == 0) {
      cnt++;
      if (dfs(e[v][i], depth + 1))
        ;
      {
        if (v == minLoc) {
          cout << v;
          exit(0);
        } else {
          cout << v << ' ';
          return true;
        }
      }
    }
  }
  if (cnt == 0) {
    int minD = depth + 1;
    for (int i = 0; i < e[v].size(); i++) {
      if (minD > d[e[v][i]]) {
        minD = d[e[v][i]];
        minLoc = e[v][i];
      }
    }
    cout << depth - minD + 1 << '\n';
    cout << v << ' ';
    return true;
  }
  return false;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    e[u].push_back(v);
    e[v].push_back(u);
  }
  dfs(1, 1);
}
