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

Copy

3 3 2
1 2
2 3
3 1

output

Copy

3
1 2 3

input

Copy

4 6 3
4 3
1 2
1 3
1 4
2 3
2 4

output

Copy

4
3 4 1 2

做这个题，我一开始想的是floyd求最小环，dijkstra求最小环，但是，到后来我发现，数据量太大，而且这个题之说找到大于K的环，便搁置了一下，太难了，后来想到了哈密尔顿，不就一个搜索题，比哈密尔顿路还简单，搜一个回路，并且记录路径，搜到重复点，就是回路，路径点大于K，就是要的答案，直接交，ojbk。

#include<iostream>
#include<queue>
#include<algorithm>
#include<set>
#include<cmath>
#include<vector>
#include<map>
#include<stack>
#include<bitset>
#include<cstdio>
#include<cstring>
//---------------------------------Sexy operation--------------------------//

#define cini(n) scanf("%d",&n)
#define cinl(n) scanf("%lld",&n)
#define cinc(n) scanf("%c",&n)
#define cins(s) scanf("%s",s)
#define coui(n) printf("%d",n)
#define couc(n) printf("%c",n)
#define coul(n) printf("%lld",n)
#define speed ios_base::sync_with_stdio(0)
#define file  freopen("input.txt","r",stdin);freopen("output.txt","w",stdout)
//-------------------------------Actual option------------------------------//

#define Swap(a,b) a^=b^=a^=b
#define Max(a,b) a>b?a:b
#define Min(a,b) a<b?a:b
#define mem(n,x) memset(n,x,sizeof(n))
#define mp(a,b) make_pair(a,b)
//--------------------------------constant----------------------------------//

#define INF  0x3f3f3f3f
#define maxn  100005
#define esp  1e-9
using namespace std;
typedef long long ll;
typedef pair<int,int> PII;
//------------------------------Dividing Line--------------------------------//
vector<int> ege[maxn],ans;
int vis[maxn];
int n,m,k,cnt;
bool dfs(int u)
{
    vis[u]=++cnt;
    ans.push_back(u);
    for(int i=0;i<ege[u].size();i++)
    {
        int v=ege[u][i];
        if(vis[v])
        {
            if(cnt-vis[v]<k) continue;
            cout<<vis[u]-vis[v]+1<<endl;
            for(int i=vis[v];i<=vis[u];i++) cout<<ans[i-1]<<' ';
            return puts(""),1;
        }
        if(dfs(v))return 1;
    }
    return 0;
}
int main()
{
    cin>>n>>m>>k;
    for(int i=0;i<m;i++)
    {
        int x,y;
        cini(x),cini(y);
        ege[x].push_back(y);
        ege[y].push_back(x);
    }
    dfs(1);
    return 0;
}