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 MAX_N = 1e5 + 5;
int n, m, k, flag;
int vis[MAX_N], deep[MAX_N];
vector<int> G[MAX_N], path;
void dfs(int pos, int step) {
  if (vis[pos]) return;
  vis[pos] = 1;
  deep[pos] = step;
  path.push_back(pos);
  int size = G[pos].size();
  for (int i = 0; i < size && !flag; i++) {
    int v = G[pos][i];
    if (vis[v] && step - deep[v] >= k) {
      printf("%d\n", step - deep[v] + 1);
      for (int j = deep[v]; j <= step; j++) {
        printf("%d%c", path[j], j == step ? '\n' : ' ');
      }
      flag = 1;
    } else {
      dfs(v, step + 1);
    }
  }
  path.pop_back();
}
int main() {
  scanf("%d%d%d", &n, &m, &k);
  int u, v;
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs(1, 0);
  return 0;
}
