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<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
struct edge{
	int s,t,nxt;
}e[200005];
int last[100005],e_cnt;
void ins(int a,int b){
	e[e_cnt]=(edge){a,b,last[a]};
	last[a]=e_cnt++;
}
int fa[100005],dep[100005];
void dfs(int x){
	for(int i=last[x];i!=-1;i=e[i].nxt)if(e[i].t!=fa[x]){
		fa[e[i].t]=x;
		dep[e[i].t]=dep[x]+1;
		dfs(e[i].t);
	}
}
vector<int> v;
int h[100005];
int p[100005],ans[100005];
int main(){
	int n;
	scanf("%d",&n);
	memset(last,-1,sizeof(last));
	for(int i=1;i<n;i++){
		int u,v;
		scanf("%d%d",&u,&v);
		ins(u,v);
		ins(v,u);
		h[u]++;
		h[v]++;
	}
	dfs(1);
	int mx=1;
	for(int i=1;i<=n;i++)if(dep[i]>dep[mx])mx=i;
	fa[mx]=0;
	dep[mx]=0;
	dfs(mx);
	for(int i=1;i<=n;i++)if(dep[i]>dep[mx])mx=i;
	while(mx)v.push_back(mx),mx=fa[mx];
	int sz=v.size();
	for(int i=1;i<v.size()-1;i++){
		sz+=h[v[i]]-2;
	}
	if(sz<n){puts("-1"); return 0;}
	int l=v.size()-1;
	for(int i=1;i<v.size()-1;i++){
		if(h[v[i]]<h[v[l-i]])break;
		else if(h[v[i]]>h[v[l-i]]){reverse(v.begin(), v.end()); break;}
	}
	p[1]=1;
	for(int i=1,lst=2;i<v.size()-1;i++){
		p[lst]=lst+h[v[i]]-2;
		for(int j=lst;j<lst+h[v[i]]-2;j++)p[j+1]=j;
		lst+=h[v[i]]-2+1;
	}
	p[n]=n;
	for(int i=1;i<=n;i++)ans[p[i]]=i;
	for(int i=1;i<=n;i++)printf("%d ",ans[i]);
	return 0;
}