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;
vector<int> graph[100010];
int used[100010];
int parent[100010];
int tim[100010];
int n, t = 0, k;
bool done = false;
void printpath(int first, int last) {
  if (first == last) {
    cout << first << " ";
    return;
  }
  printpath(first, parent[last]);
  cout << last << " ";
  return;
}
void Dfs(int i) {
  t++;
  used[i] = true;
  tim[i] = t;
  for (int h = 0; h < graph[i].size(); h++) {
    int j = graph[i][h];
    if (!used[j]) {
      parent[j] = i;
      Dfs(j);
      if (done) return;
    } else if (used[j] && parent[j] != i && tim[i] - tim[j] >= k) {
      cout << tim[i] - tim[j] + 1 << endl;
      printpath(j, i);
      done = true;
      return;
    }
  }
}
int main() {
  int m;
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    graph[x].push_back(y);
    graph[y].push_back(x);
  }
  for (int i = 0; i < n && !done; i++) {
    if (!used[i]) {
      Dfs(i);
    }
  }
  return 0;
}
