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
//ΔARC 095F
#include<iostream>
#include<cstdio>
#include<fstream>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<cmath>
#include<cstring>
#include<cstdlib>
using namespace std;
typedef long long LL;
typedef double DB;
const int N = 111111;
int n,d[N],l,f[N],g[N],ans1[N],ans2[N];
vector<int> v[N];
void dfs(int u,int fa){
	d[u]=d[fa]+1;
	int i=v[u].size();
	while(i--){
		if(v[u][i]!=fa)
			dfs(v[u][i],u);
	}
}
int main()
{
	int i,x,y,s;
	scanf("%d",&n);
	for(i=1;i<n;i=i+1){
		scanf("%d%d",&x,&y);
		v[x].push_back(y);
		v[y].push_back(x);
	}
	dfs(1,0);
	y=1;
	for(i=1;i<=n;i=i+1)
		if(d[i]>y)
			x=i,y=d[i];
	dfs(x,0);
	y=1;
	for(i=1;i<=n;i=i+1)
		if(d[i]>y)
			x=i,y=d[i];
	while(1){
		f[++l]=x,g[l]=v[x].size()-1;
		if(d[x]==1)
			break;
		i=v[x].size();
		while(i--){
			if(d[v[x][i]]<d[x]){
				x=v[x][i];
				break;
			}
		}
	}
	s=n;
	for(i=1;i<=l;i=i+1){
		x=f[i];
		y=v[x].size();
		while(y--){
			if(d[v[x][y]])
				d[v[x][y]]=0,s--;
		}
	}
	if(s){
		cout<<-1;
		return 0;
	}
	g[1]=1,g[l]=1;
	x=0;
	ans1[1]=++x;
	for(i=2;i<=l;i=i+1){
		ans1[x+g[i]]=x+1;
		for(y=x+1;y<x+g[i];y=y+1)
			ans1[y]=y+1;
		x+=g[i];
	}
	x=0;
	ans2[1]=++x;
	for(i=l-1;i;i=i-1){
		ans2[x+g[i]]=x+1;
		for(y=x+1;y<x+g[i];y=y+1)
			ans2[y]=y+1;
		x+=g[i];
	}
	x=0;
	for(i=1;i<=n;i=i+1){
		if(!x){
			if(ans1[i]<ans2[i])
				x=1;
			if(ans1[i]>ans2[i])
				x=2;
		}
		if(x<2)
			printf("%d ",ans1[i]);
		else
			printf("%d ",ans2[i]);
	}
	return 0;
}