Takahashi has an ability to generate a tree using a permutation (p_1,p_2,...,p_n) of (1,2,...,n), in the following process:

First, prepare Vertex 1, Vertex 2, ..., Vertex N. For each i=1,2,...,n, perform the following operation:

* If p_i = 1, do nothing.
* If p_i \neq 1, let j' be the largest j such that p_j < p_i. Span an edge between Vertex i and Vertex j'.



Takahashi is trying to make his favorite tree with this ability. His favorite tree has n vertices from Vertex 1 through Vertex n, and its i-th edge connects Vertex v_i and w_i. Determine if he can make a tree isomorphic to his favorite tree by using a proper permutation. If he can do so, find the lexicographically smallest such permutation.

Constraints

* 2 \leq n \leq 10^5
* 1 \leq v_i, w_i \leq n
* The given graph is a tree.

Input

Input is given from Standard Input in the following format:


n
v_1 w_1
v_2 w_2
:
v_{n-1} w_{n-1}


Output

If there is no permutation that can generate a tree isomorphic to Takahashi's favorite tree, print `-1`. If it exists, print the lexicographically smallest such permutation, with spaces in between.

Examples

Input

6
1 2
1 3
1 4
1 5
5 6


Output

1 2 4 5 3 6


Input

6
1 2
2 3
3 4
1 5
5 6


Output

1 2 3 4 5 6


Input

15
1 2
1 3
2 4
2 5
3 6
3 7
4 8
4 9
5 10
5 11
6 12
6 13
7 14
7 15


Output

-1
#include<bits/stdc++.h>
using namespace std;
#define N 100011
typedef struct{
	int to,nxt;
}edge; 
edge e[N<<1];
queue<int> q;
int cnt,head[N<<1],dep[N],fa[N],siz[N],vis[N],ans,a,b,n,st,ed,sta[N],top;
int p1[N],p2[N],np[N],val[N];
void add(int a,int b){
	cnt++;
	e[cnt].to=b;
	e[cnt].nxt=head[a];
	head[a]=cnt;
}
void bfs(int s){
	dep[s]=1;
	q.push(s);
	while(!q.empty()){
		int o=q.front();q.pop();
		for(int i=head[o];i!=-1;i=e[i].nxt){
			int v=e[i].to;
			if(!dep[v]){
				dep[v]=dep[o]+1;
				q.push(v);
			}
		}
	}
}
void dfs(int u,int f){
	fa[u]=f;
	for(int i=head[u];i!=-1;i=e[i].nxt){
		int v=e[i].to;
		if(v==f) continue;
		dfs(v,u);
	} 
}
int main(){
	memset(head,-1,sizeof(head));
	scanf("%d",&n);
	for(int i=1;i<=n-1;i++){
		scanf("%d%d",&a,&b);
		add(a,b);add(b,a);
	}
	bfs(1);int mx=0;
	for(int i=1;i<=n;i++)
	  if(mx<dep[i]){
	  	mx=dep[i];
	  	st=i;
	  }
	memset(dep,0,sizeof(dep));
	bfs(st);mx=0;
	for(int i=1;i<=n;i++)
	  if(mx<dep[i]){
	  	mx=dep[i];ed=i;
	  }
	dfs(st,0);
	int x=ed;
	while(x){
		vis[x]=2;
		for(int i=head[x];i!=-1;i=e[i].nxt){
			int v=e[i].to;
			if(v==fa[x]) continue;
			if(!vis[v]) vis[v]=1;
		}
		x=fa[x];
	}
	for(int i=1;i<=n;i++)
	  if(!vis[i]){
	  	printf("-1\n");
	  	return 0;
	  }
	for(int i=1;i<=n;i++){
		if(vis[i]==1){
			val[fa[i]]++;
		}
	}
	x=ed;
	while(x){
		sta[++top]=val[x]+1;
		x=fa[x];
	}
	int ccnt=0;
	for(int i=1;i<=top;i++){
		np[ccnt+1]=ccnt+sta[i];
		for(int j=1;j<=sta[i]-1;j++)
		  np[ccnt+1+j]=ccnt+j;
		ccnt+=sta[i];
	}
	for(int i=1;i<=n;i++) p1[np[i]]=i;
	ccnt=0;
	for(int i=top;i>=1;i--){
		np[ccnt+1]=ccnt+sta[i];
		for(int j=1;j<=sta[i]-1;j++)
		  np[ccnt+1+j]=ccnt+j;
		ccnt+=sta[i];
	}
	for(int i=1;i<=n;i++) p2[np[i]]=i;
	for(int i=1;i<=n;i++)
	  if(p1[i]>p2[i]){
	  	for(int j=1;j<=n;j++)
	  	  printf("%d ",p2[j]);
	  	printf("\n");
	  	return 0;
	  }else
	  if(p1[i]<p2[i]){
	  	for(int j=1;j<=n;j++)
	  	  printf("%d ",p1[j]);
	  	printf("\n");
	  	return 0;
	  }
	for(int i=1;i<=n;i++)
	  printf("%d ",p1[i]);
	printf("\n");
}