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;
vector<vector<int> > g;
vector<int> a, q;
vector<char> cl;
vector<int> p;
map<int, pair<int, int> > y;
map<int, pair<int, int> >::iterator it;
int cycle_st, cycle_end;
long long i, j, n, m, e, k, ot, x, i1, i2, i3, i4, l, r;
bool dfs(int v) {
  cl[v] = 1;
  for (size_t i = 0; i < g[v].size(); ++i) {
    int to = g[v][i];
    if (cl[to] == 0 && p[v] != to) {
      p[to] = v;
      y[v].first = to;
      y[to].second = y[v].second + 1;
      if (dfs(to)) return true;
    } else if (cl[to] == 1 && p[v] != to &&
               y[v].second - y[to].second + 1 >= k + 1) {
      cycle_end = v;
      cycle_st = to;
      cout << y[v].second - y[to].second + 1 << endl;
      while (y[to].first != v) {
        cout << to << " ";
        to = y[to].first;
      }
      cout << to << " " << v << " ";
      exit(0);
      return true;
    }
  }
  cl[v] = 2;
  return false;
}
int main() {
  vector<vector<int> > w;
  vector<int> d;
  cin >> n >> m >> k;
  for (i = 0; i < n + 2; i++) {
    g.push_back(a);
    p.push_back(-1);
    cl.push_back(0);
  }
  for (i = 0; i < m; i++) {
    cin >> l >> r;
    g[l].push_back(r);
    g[r].push_back(l);
  }
  for (int i = 0; i < n; ++i) {
    y[i].second = 1;
    if (dfs(i)) break;
  }
  return 0;
}
