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One day Bob got a letter in an envelope. Bob knows that when Berland’s post officers send a letter directly from city «A» to city «B», they stamp it with «A B», or «B A». Unfortunately, often it is impossible to send a letter directly from the city of the sender to the city of the receiver, that’s why the letter is sent via some intermediate cities. Post officers never send a letter in such a way that the route of this letter contains some city more than once. Bob is sure that the post officers stamp the letters accurately.

There are n stamps on the envelope of Bob’s letter. He understands that the possible routes of this letter are only two. But the stamps are numerous, and Bob can’t determine himself none of these routes. That’s why he asks you to help him. Find one of the possible routes of the letter.

Input

The first line contains integer n (1 ≤ n ≤ 105) — amount of mail stamps on the envelope. Then there follow n lines with two integers each — description of the stamps. Each stamp is described with indexes of the cities between which a letter is sent. The indexes of cities are integers from 1 to 109. Indexes of all the cities are different. Every time the letter is sent from one city to another, exactly one stamp is put on the envelope. It is guaranteed that the given stamps correspond to some valid route from some city to some other city.

Output

Output n + 1 numbers — indexes of cities in one of the two possible routes of the letter.

Examples

Input 2 1 100 100 2 Output 2 100 1 Input 3 3 1 100 2 3 2 Output 100 2 3 1 题意很简单，就是给你一张图，输出一条路径，里面包括所有的点一次且仅有一次。 点不是连续的，我们需要离散化一下。之后就dfs找路径就可以了。 代码如下：

#include
   
    #define ll long longusing namespace std;const int maxx=2e5+100;struct node{
   	int x,y;}p[maxx];struct edge{
   	int to,next;}e[maxx<<1];int head[maxx<<1];int in[maxx];int vis[maxx],a[maxx];int n,tot,ans[maxx];/*-----------事前准备-----------*/inline void init(){
   	memset(head,-1,sizeof(head));	tot=0;}inline void add(int u,int v){
   	e[tot].to=v,e[tot].next=head[u],head[u]=tot++;}/*-----------dfs-----------*/inline void dfs(int u,int cnt){
   	vis[u]=1;	for(int i=head[u];i!=-1;i=e[i].next)	{
   		int to=e[i].to;		if(vis[to]) continue;//走过的点不重复走。		ans[cnt]=a[to];		dfs(to,cnt+1);	}}int main(){
   	int cnt=0,x,y;	scanf("%d",&n);init();	for(int i=1;i<=n;i++)	{
   		scanf("%d%d",&x,&y);		p[i].x=x;p[i].y=y;		a[++cnt]=x,a[++cnt]=y;	}	sort(a+1,a+1+cnt);	int len=unique(a+1,a+1+cnt)-a-1;	for(int i=1;i<=n;i++)	{
   		p[i].x=lower_bound(a+1,a+1+len,p[i].x)-a;		p[i].y=lower_bound(a+1,a+1+len,p[i].y)-a;//离散化之后连边		add(p[i].x,p[i].y);		add(p[i].y,p[i].x);		in[p[i].x]++;in[p[i].y]++;	}	int i;	for(i=1;i<=len;i++)	{
   		if(in[i]==1) break;	}	ans[1]=a[i];	dfs(i,2);	for(int i=1;i<=len;i++) cout<
    
     <<" ";	cout<
    
   

努力加油a啊，(^o)/~

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