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 N = 1e5 + 5;
int low[N];
vector<int> adj[N];
int main() {
  int n, m, k;
  while (scanf("%d%d%d", &n, &m, &k) != EOF) {
    for (int i = 1; i <= n; i++) adj[i].clear();
    memset(low, 0, sizeof(low));
    for (int i = 0; i < m; i++) {
      int a, b;
      scanf("%d%d", &a, &b);
      adj[a].push_back(b);
      adj[b].push_back(a);
    }
    vector<int> res;
    res.push_back(1);
    low[1] = 1;
    while (1) {
      int u = res.back(), mi = (1 << 30), nxt;
      for (int i = 0; i < (int)adj[u].size(); i++) {
        int v = adj[u][i];
        if (mi > low[v]) {
          mi = low[v];
          nxt = v;
        }
      }
      if (mi == 0) {
        res.push_back(nxt);
        low[nxt] = (int)res.size();
      } else {
        int len = (int)res.size();
        printf("%d\n", len - mi + 1);
        for (int i = mi - 1; i < len; i++)
          printf("%d%c", res[i], i == len - 1 ? '\n' : ' ');
        break;
      }
    }
  }
  return 0;
}
