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 x,y,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 NO(){
	puts("-1");
	exit(0);
}
int main(){
	scanf("%d",&n);
	for (int i=1;i<n;i++){
		scanf("%d%d",&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)NO();
			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())NO();
	for (int i=0;i<n;i++)printf("%d ",ans[i]);
	return 0;
}
