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 int maxN = 100500;
const int INF = (int)1e9;
const int MOD = 7 + (int)1e9;
const double eps = 1e-9;
vector<int> G[maxN];
int mark[maxN];
vector<int> cur;
int k;
int ii;
void dfs(int v) {
  if (mark[v]) return;
  mark[v] = 1;
  int f = 0;
  cur.push_back(v);
  for (int i = 0; i < G[v].size(); i++) {
    int u = G[v][i];
    if (u == ii) {
      f = 1;
    }
  }
  if (f && cur.size() > k) {
    cout << cur.size() << endl;
    for (int i = 0; i < cur.size(); i++) {
      cout << cur[i] << " ";
    }
    cout << endl;
    exit(0);
  }
  for (int i = 0; i < G[v].size(); i++) {
    int u = G[v][i];
    dfs(u);
  }
  cur.pop_back();
}
int main(void) {
  srand(time(NULL));
  int n, m;
  cin >> n >> m >> k;
  while (m--) {
    int v, u;
    cin >> v >> u;
    G[v].push_back(u);
    G[u].push_back(v);
  }
  for (int i = 0; i < 100; i++) {
    cur.clear();
    for (int i = 0; i < maxN; i++) {
      mark[i] = 0;
    }
    ii = rand() % n + 1;
    dfs(ii);
  }
  return 0;
}
