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<stdio.h>
int head[131072],last[262144],to[262144],cnt=0;
void add(int u,int v)
{
	cnt++;
	last[cnt]=head[u];
	head[u]=cnt;
	to[cnt]=v;
}
int deg[131072];
int a[131072],num=0;
void dfs(int u,int f)
{
	a[num]=deg[u]-2;
	num++;
	for(int i=head[u];i;i=last[i])
	{
		int v=to[i];
		if(v==f)
		{
			continue;
		}
		dfs(v,u);
	}
}
int now=0;
int ans[131072];
void add(int val)
{
	ans[now+val+1]=now+1;
	for(int i=now+2;i<=now+val+1;i++)
	{
		ans[i-1]=i;
	}
	now=now+val+1;
}
int deg2[131072];
int u[131072],v[131072];
int main()
{
	int n;
	scanf("%d",&n);
	for(int i=1;i<n;i++)
	{
		scanf("%d%d",&u[i],&v[i]);
		deg[u[i]]++;
		deg[v[i]]++;
	}
	if(n==2)
	{
		printf("1 2\n");
		return 0;
	}
	for(int i=1;i<n;i++)
	{
		if(deg[u[i]]>1&&deg[v[i]]>1)
		{
			add(u[i],v[i]);
			add(v[i],u[i]);
			deg2[u[i]]++;
			deg2[v[i]]++;
		}
	}
	for(int i=1;i<=n;i++)
	{
		if(deg2[i]>2)
		{
			printf("-1\n");
			return 0;
		}
	}
	for(int i=1;i<=n;i++)
	{
		if(deg[i]>1&&deg2[i]<=1)
		{
			dfs(i,0);
			break;
		}
	}
	bool ok=true;
	for(int i=0;i<num;i++)
	{
		if(a[i]!=a[num-i-1])
		{
			ok=(a[i]<a[num-i-1]);
			break;
		}
	}
	add(0);
	if(ok)
	{
		for(int i=0;i<num;i++)
		{
			add(a[i]);
		}
	}
	else
	{
		for(int i=num-1;i>=0;i--)
		{
			add(a[i]);
		}
	}
	add(0);
	for(int i=1;i<=n;i++)
	{
		printf("%d",ans[i]);
		if(i==n)
		{
			putchar('\n');
		}
		else
		{
			putchar(' ');
		}
	}
	return 0;
}