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(100001);
bool flag[100001];
int a[100001];
int n, m, k, p;
bool ans;
void enter(int v) {
  a[p] = v;
  p++;
}
void exit() {
  if (!ans) p--;
}
void DFS(int v, int s) {
  enter(v);
  flag[v] = true;
  for (__typeof((g[v]).begin()) it = (g[v]).begin(); it != (g[v]).end(); it++) {
    if (ans) return;
    if (*it == 1) {
      if (s > k) {
        ans = true;
        return;
      }
    } else {
      if (!flag[*it]) DFS(*it, s + 1);
    }
  }
  exit();
}
int main() {
  int i, x, y;
  cin >> n >> m >> k;
  for ((i) = 0; (i) < int(m); (i)++) {
    scanf("%d%d", &x, &y);
    g[x].push_back(y);
    g[y].push_back(x);
  }
  DFS(1, 1);
  cout << p << endl;
  for ((i) = 0; (i) < int(p); (i)++) printf("%d ", a[i]);
}
