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 rep(i,n) for(int i=0;i<(n);i++)
#define REP(i,n) for(int i=1;i<=(n);i++)
#define mp make_pair
#define pb push_back
#define fst first
#define snd second
typedef long long ll;
typedef pair<int,int> pii;

const int maxn=100005;
int n,ql,qr,len;
vector<int>edge[maxn];
int dist[maxn],que[maxn],par[maxn],path[maxn];
int ans1[maxn],ans2[maxn];

void bfs(int s){
	REP(i,n)dist[i]=n;
	dist[s]=0;
	ql=1;qr=0;
	que[++qr]=s;
	while(ql<=qr){
		int x=que[ql++];
		rep(i,int(edge[x].size())){
			int y=edge[x][i];
			if(dist[y]==n){
				par[y]=x;
				dist[y]=dist[x]+1;
				que[++qr]=y;
			}
		}
	}
}

int main(){
	scanf("%d",&n);
	REP(i,n-1){
		int u,v;
		scanf("%d%d",&u,&v);
		edge[u].pb(v);
		edge[v].pb(u);
	}
	bfs(1);
	int s=1,t=1;
	REP(i,n)if(dist[i]>dist[s])s=i;
	bfs(s);
	REP(i,n)if(dist[i]>dist[t])t=i;
	while(t!=s)path[++len]=t,t=par[t];
	path[++len]=s;
	int now=0;
	REP(i,len){
		int cur=int(edge[path[i]].size());
		if(i>1)cur--;
		if(i<len)cur--;
		REP(j,cur)ans1[now+j]=now+1+j;
		ans1[now+1+cur]=now+1;
		now+=cur+1;
	}
	if(now!=n){
		puts("-1");
		return 0;
	}
	now=0;
	for(int i=len;i>=1;i--){
		int cur=int(edge[path[i]].size());
		if(i>1)cur--;
		if(i<len)cur--;
		REP(j,cur)ans2[now+j]=now+1+j;
		ans2[now+1+cur]=now+1;
		now+=cur+1;
	}
	bool ok=false;
	REP(i,n)if(ans1[i]!=ans2[i]){
		ok=(ans2[i]<ans1[i]);
		break;
	}
	if(ok)REP(i,n)ans1[i]=ans2[i];
	REP(i,n)printf("%d%c",ans1[i],i==n?'\n':' ');
	return 0;
}