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>
#define rep(i,n)for(int i=0;i<(n);i++)
using namespace std;

vector<int>E[200000];
int d[200000];

void dfs(int v){
	for(int u:E[v]){
		if(d[u]==-1){
			d[u]=d[v]+1;
			dfs(u);
		}
	}
}
vector<int>ans;
int cnt=1;

bool dfs2(int v,int p){
	ans.push_back(cnt++);
	vector<int>c;
	for(int u:E[v]){
		if(u==p)continue;
		c.push_back(u);
	}
	if(c.size()==0)return 1;
	int id=-1;
	rep(i,c.size()){
		int u=c[i];
		if(E[u].size()!=1){
			if(id!=-1)return 0;
			id=u;
		}
		else{
			if(i==c.size()-1&&id==-1){
				ans.push_back(cnt++);
			}
			else{
				int d=ans.back();ans.pop_back();
				ans.push_back(cnt++);
				ans.push_back(d);
			}
		}
	}
	if(id!=-1)return dfs2(id,v);
	return 1;
}
int main(){
	int n;scanf("%d",&n);
	rep(i,n-1){
		int a,b;scanf("%d%d",&a,&b);a--;b--;
		E[a].push_back(b);
		E[b].push_back(a);
	}
	memset(d,-1,sizeof(d));
	d[0]=0;dfs(0);
	int u=max_element(d,d+n)-d;
	memset(d,-1,sizeof(d));
	d[u]=0;dfs(u);
	int v=max_element(d,d+n)-d;
	if(!dfs2(u,-1)){
		puts("-1");return 0;
	}
	vector<int>res1=ans;
	ans.clear();
	cnt=1;
	dfs2(v,-1);
	if(res1<ans)ans=res1;
	rep(i,ans.size()){
		if(i)printf(" ");
		printf("%d",ans[i]);
	}
	puts("");
}