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<cstdio>
using namespace std;
#define N 205000
int head[N],cnt,dep[N],l,r,in[N],st[N],n,a,b,ct,st2[N],fg,as1[N],as2[N],is[N];
struct edge{int t,next;}ed[N*2];
void adde(int f,int t){ed[++cnt]=(edge){t,head[f]};head[f]=cnt;ed[++cnt]=(edge){f,head[t]};head[t]=cnt;in[f]++,in[t]++;}
void dfs(int u,int fa){dep[u]=dep[fa]+1;for(int i=head[u];i;i=ed[i].next)if(ed[i].t!=fa)dfs(ed[i].t,u);}
void dfs2(int u,int t,int fa){if(!fg)st[++ct]=u;if(u==t)fg=1;for(int i=head[u];i;i=ed[i].next)if(ed[i].t!=fa)dfs2(ed[i].t,t,u);if(!fg)ct--;}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<n;i++)scanf("%d%d",&a,&b),adde(a,b);
	dfs(1,0);int mx=0,vl;for(int i=1;i<=n;i++)if(dep[i]>mx)mx=dep[i],vl=i;
	dfs(vl,0);mx=0;int vl2=0;for(int i=1;i<=n;i++)if(dep[i]>mx)mx=dep[i],vl2=i;
	dfs2(vl,vl2,0);
	for(int i=1;i<=ct;i++)is[st[i]]=1;
	for(int i=1;i<=n;i++)if(!is[i]&&in[i]>1){printf("-1\n");return 0;}
	for(int i=1;i<=ct;i++)st2[i]=in[st[i]]-2+(in[st[i]]==1);
	int lb=0;
	for(int i=1;i<=ct;i++)
	{
		for(int j=1;j<=st2[i];j++)as1[lb+j]=lb+j+1;
		as1[lb+st2[i]+1]=lb+1;
		lb+=st2[i]+1;
	}
	lb=0;
	for(int i=ct;i>=1;i--)
	{
		for(int j=1;j<=st2[i];j++)as2[lb+j]=lb+j+1;
		as2[lb+st2[i]+1]=lb+1;
		lb+=st2[i]+1;
	}
	int fg2=0;
	for(int i=1;i<=n;i++)if(as1[i]<as2[i]){fg2=1;break;}else if(as2[i]<as1[i])break;
	if(!fg2)for(int i=1;i<=n;i++)as1[i]=as2[i];
	for(int i=1;i<=n;i++)printf("%d ",as1[i]);
}