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;
int n, m, k, p[100001];
vector<int> g[100001];
int was[100001];
int num = 1;
bool found = 0;
int start, cnt;
void dfs(int v) {
  if (!found) {
    was[v] = num++;
    bool f = 0;
    for (int i = 0; i < g[v].size(); i++) {
      int to = g[v][i];
      if (!was[to]) {
        p[to] = v;
        dfs(to);
        f = 1;
        break;
      }
    }
    if (found) {
      return;
    }
    if (!f) {
      found = 1;
      int min = g[v][0];
      for (int i = 1; i < g[v].size(); i++) {
        int to = g[v][i];
        if (was[to] < was[min]) {
          min = to;
        }
      }
      p[min] = v;
      start = v;
      cnt = was[v] - was[min] + 1;
    }
  }
}
int main() {
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  memset(p, -1, sizeof(p));
  dfs(1);
  cout << cnt << endl;
  int v = start;
  do {
    cout << v << ' ';
    v = p[v];
  } while (v != start);
  return 0;
}
