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<iostream>
#include<cstdio>
#include<vector>
#include<algorithm>
#define FOR(i,a,b) for(int i=a;i<=b;i++) 
using namespace std;
const int N=100100;
int n,S,T,u,v,m,l,r,sz[N],fa[N],d[N],a[N],b[N],p[N],q[N];
vector<int>ve[N];
void add(int u,int v){ve[u].push_back(v);}
void dfs(int u,int&w){
	if(d[u]>d[w]) w=u;
	for(int v:ve[u])if(v!=fa[u]) fa[v]=u,d[v]=d[u]+1,dfs(v,w);
}
void work(int*p){
	l=u=0;
	FOR(i,1,m){
		p[l+sz[a[i]]+1]=++u;
		FOR(j,1,sz[a[i]]) p[++l]=++u;l++;
	}
}
int main(){
	//freopen("1.in","r",stdin);
	scanf("%d",&n);
	FOR(i,2,n) scanf("%d%d",&u,&v),add(u,v),add(v,u);
	dfs(1,T);fa[T]=d[T]=0;dfs(T,S);
	for(u=S;u;u=fa[u]) a[++m]=u,b[u]=1;
	FOR(i,1,n)if(!b[i] && !b[fa[i]]) return puts("-1"),0;
	FOR(i,1,n)if(!b[i]) sz[fa[i]]++;
	work(p);reverse(a+1,a+m+1);work(q);
	FOR(i,1,n)if(p[i]!=q[i]){
		if(p[i]>q[i]) swap(p,q);
		break;
	}
	FOR(i,1,n) cout<<p[i]<<' ';cout<<'\n';
}