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 = 100000 + 5;
vector<int> G[maxn];
int fa[maxn];
int vis[maxn];
int chuo[maxn];
int k, tag;
void dfs(int x, int cnt) {
  for (int i = 0; i < G[x].size(); i++) {
    if (tag == 1) return;
    int u = G[x][i];
    if (!vis[u]) {
      vis[u] = 1;
      fa[u] = x;
      chuo[u] = cnt + 1;
      dfs(u, cnt + 1);
      vis[u] = 0;
    } else if (chuo[x] - chuo[u] >= k) {
      cout << chuo[x] - chuo[u] + 1 << endl;
      int tem = x;
      cout << x << ' ';
      while (fa[tem] != u) {
        cout << fa[tem] << ' ';
        tem = fa[tem];
      }
      cout << u << endl;
      tag = 1;
      return;
    }
  }
}
int main() {
  int n, m;
  while (cin >> n >> m >> k) {
    for (int i = 1; i <= n; i++) G[i].clear();
    while (m--) {
      int x, y;
      cin >> x >> y;
      G[x].push_back(y);
      G[y].push_back(x);
    }
    memset(vis, 0, sizeof(vis));
    vis[1] = 1;
    chuo[1] = 0;
    tag = 0;
    dfs(1, 0);
  }
  return 0;
}
