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>
#pragma comment(linker, "/STACK:16777216")
using namespace std;
const double eps = 1e-9;
const double pi = acos(-1.0);
const int maxn = (int)1e5 + 10;
int k;
vector<int> g[maxn], cur;
int used[maxn];
void dfs(int v) {
  for (int i = 0; i < (int)(g[v]).size(); i++) {
    int to = g[v][i];
    if (used[to] != -1 && (int)(cur).size() - used[to] > k) {
      cout << (int)(cur).size() - used[to] << endl;
      for (int i = used[to]; i < (int)(cur).size(); i++) {
        cout << cur[i] + 1 << " ";
      }
      exit(0);
    } else if (used[to] == -1) {
      used[to] = (int)(cur).size();
      cur.push_back(to);
      dfs(to);
      used[to] = -1;
      cur.pop_back();
    }
  }
}
int main() {
  int n, m;
  cin >> n >> m >> k;
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    --a;
    --b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  memset(used, -1, sizeof(used));
  for (int i = 0; i < n; i++) {
    used[i] = 0;
    cur.push_back(i);
    dfs(i);
    used[i] = -1;
    cur.pop_back();
  }
  return 0;
}
