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 tmax = (int)1e5 + 100;
vector<int> graph[tmax];
vector<pair<int, pair<int, int> > > cycle;
int visited[tmax], parent[tmax];
int n, m, k;
void dfs(int node, int cc) {
  visited[node] = cc;
  for (int i = 0; i < graph[node].size(); i++) {
    int no = graph[node][i];
    if (visited[no] == -1) {
      dfs(no, cc + 1);
      parent[no] = node;
    } else if (abs(visited[node] - visited[no]) >= k)
      cycle.push_back(
          make_pair(node, make_pair(no, visited[node] - visited[no])));
  }
}
void rec(int node, int fim) {
  printf("%d ", node);
  if (node == fim) return;
  rec(parent[node], fim);
}
int main() {
  scanf("%d %d %d", &n, &m, &k);
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    graph[a].push_back(b);
    graph[b].push_back(a);
  }
  set<int> used;
  for (int i = 0; i < 1000; i++) {
    int newpos = rand() % n + 1;
    if (used.find(newpos) != used.end()) continue;
    used.insert(newpos);
    for (int j = 0; j <= n; j++) visited[j] = -1, parent[j] = -1;
    cycle.clear();
    dfs(newpos, 1);
    if (cycle.size()) {
      printf("%d\n", cycle[0].second.second + 1);
      rec(cycle[0].first, cycle[0].second.first);
      printf("\n");
      return 0;
    }
  }
  return 0;
}
