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 MX = 1e5 + 10, MD = 1e9 + 7;
long long n, m, a[MX], ans, tmp, k, h[MX], p[MX];
vector<int> adj[MX];
bool mark[MX];
void dfs(int u) {
  mark[u] = 1;
  for (auto v : adj[u]) {
    if (!mark[v]) {
      p[v] = u;
      h[v] = h[u] + 1;
      dfs(v);
    } else {
      if (h[u] - h[v] + 1 >= k + 1) {
        printf("%d\n", h[u] - h[v] + 1);
        tmp = u;
        while (v != tmp) {
          printf("%lld ", tmp);
          tmp = p[tmp];
        }
        printf("%lld ", tmp);
        exit(0);
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  scanf("%d%d%d", &n, &m, &k);
  int u, v;
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1);
}
