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;

const int N=100100;
int n,clk;
vector<int>vec,ans,g[N];

void dfs(int x,int fa=-1){
	int dfn=++clk,sz=g[x].size();
	if(sz==1&&fa!=-1){vec.push_back(clk);return;}
	int id=0;
	for(int j:g[x])if(j!=fa){
		if((int)g[j].size()>1){
			if(id&&(int)g[id].size()>1)return;
			if(id)dfs(id,x);id=j;
		}else if(!id)id=j;else dfs(j,x);
	}
	vec.push_back(dfn);
	dfs(id,x);
}

void gofail(){
	cout<<-1<<'\n';
	exit(0);
}

int main(){
	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin>>n;
	for(int i=1;i<n;++i){
		int x,y;cin>>x>>y;
		g[x].push_back(y);g[y].push_back(x);
	}
	int cnt=0;
	for(int i=1;i<=n;++i){
		int tmp=0;
		for(int j:g[i])if((int)g[j].size()>1)++tmp;
		if(tmp<=1&&(n==2||(int)g[i].size()>1)){
			++cnt;if(cnt>=3)gofail();
			vec.clear();clk=0;
			dfs(i);
			if((int)vec.size()==n){
				if(ans.empty())ans=vec;
				else ans=min(ans,vec);
			}
			for(int j:g[i])if((int)g[j].size()==1){
				vec.clear();clk=0;
				dfs(j);
				if((int)vec.size()==n){
					if(ans.empty())ans=vec;
					else ans=min(ans,vec);
				}
				break;
			}
		}
	}
	if(ans.empty())gofail();
	for(int i=0;i<n;++i)cout<<ans[i]<<" \n"[i==n-1];
	return 0;
}

