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>
#include<algorithm>
#include<cstring>
using namespace std;
struct Edge
{
	int to;
	int nxt;
}e[200005];
int n,m,edgenum,head[100005],dep[100005],pa[100005],d[100005],a[100005],top;
int f[100005],g[100005];
void add(int u,int v)
{
	e[++edgenum].to=v;
	e[edgenum].nxt=head[u];
	head[u]=edgenum;
}
void dfs(int node)
{
	dep[node]=dep[pa[node]]+1;
	for(int hd=head[node];hd;hd=e[hd].nxt)
	{
		int to=e[hd].to;
		if(to==pa[node])continue;
		pa[to]=node;
		dfs(to);
	}
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<n;i++)
	{
		int u,v;
		scanf("%d%d",&u,&v);
		add(u,v);
		add(v,u);
		d[u]++,d[v]++;
	}
	dfs(1);
	int u=0,v=0;
	for(int i=1;i<=n;i++)
	{
		if(dep[i]>dep[u])
		  u=i;
	}
	pa[u]=0;
	dfs(u);
	for(int i=1;i<=n;i++)
	{
		if(dep[i]>dep[v])
		  v=i;
	}
	while(v!=u)
	{
		a[++top]=d[v];
		v=pa[v];
	}
	top++;
	int now=0;
	for(int i=1;i<=top;i++)
	{
		int pre=now;
		for(int j=1;j<=a[i]-1-(i!=1);j++)
		{
			now++;
			f[now]=now+1;
		}
		f[++now]=pre+1;
	}
	if(now!=n)
	{
		printf("-1\n");
		return 0;
	}
	now=0;
	for(int i=top;i>=1;i--)
	{
		int pre=now;
		for(int j=1;j<=a[i]-1-(i!=1);j++)
		{
			now++;
			g[now]=now+1;
		}
		g[++now]=pre+1;
	}
	for(int i=1;i<=n;i++)
	{
		if(f[i]<g[i])
		{
			for(int j=1;j<=n;j++)printf("%d ",f[j]);
			printf("\n");
			return 0;
		}
		if(f[i]>g[i])
		{
			for(int j=1;j<=n;j++)printf("%d ",g[j]);
			printf("\n");
			return 0;
		}
	}
	for(int i=1;i<=n;i++)printf("%d ",f[i]);
	printf("\n");
	return 0;
}