#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct bian
{
	int nxt,to;
}bi[200020];
int n,head[100010],num,d[100010],ans[100010],cur[100010],p;
bool leaf[100010];
inline void add(int from,int to){bi[++num]=bian{head[from],to};head[from]=num;}
void dfs(int v,int fa)
{
	int cnt=0,s=0;
	for(int i=head[v],u;i;i=bi[i].nxt)
	{
		u=bi[i].to;
		if(u==fa)continue;
		if(leaf[u])++cnt;
		else s=u;
	}
	if(!s)
	{
		for(int i=1;i<=cnt-1;++i)cur[p+i]=p+i+1;
		cur[p+cnt]=p+1;
		cur[p+cnt+1]=p+cnt+1;
	}
	else
	{
		for(int i=1;i<=cnt;++i)cur[p+i]=p+i+1;
		cur[p+cnt+1]=p+1;
		p+=cnt+1;
		dfs(s,v);
	}
}
void check(int v)
{
//	printf("%d\n",v);
	p=0;dfs(v,0);
//	for(int i=1;i<=n;++i)printf("%d ",cur[i]);printf("\n");
	bool ok;
	if(!ans[1])ok=1;
	else
	{
		for(int i=1;i<=n;++i)
			if(cur[i]!=ans[i])
			{
				ok=cur[i]<ans[i];
				break;
			}
	}
	if(ok)memcpy(ans,cur,sizeof(cur));
}
int main()
{
	scanf("%d",&n);
	if(n==2)return printf("1 2"),0;
	int x,y;
	for(int i=1;i<n;++i)
	{
		scanf("%d %d",&x,&y);
		add(x,y),add(y,x);++d[x],++d[y];
	}
	for(int i=1;i<=n;++i)if(d[i]==1)leaf[i]=1;
	for(int i=1;i<=n;++i)if(leaf[i])--d[bi[head[i]].to];
	for(int i=1;i<=n;++i)
		if(!leaf[i]&&d[i]>2)return printf("-1"),0;
	for(int i=1;i<=n;++i)
		if(!leaf[i]&&d[i]<=1)
		{
			check(i);
			for(int j=head[i];j;j=bi[j].nxt)
				if(leaf[bi[j].to]){check(bi[j].to);break;}
		}
	for(int i=1;i<=n;++i)printf("%d ",ans[i]);
	return 0;
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

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

### Response

```cpp
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct bian
{
	int nxt,to;
}bi[200020];
int n,head[100010],num,d[100010],ans[100010],cur[100010],p;
bool leaf[100010];
inline void add(int from,int to){bi[++num]=bian{head[from],to};head[from]=num;}
void dfs(int v,int fa)
{
	int cnt=0,s=0;
	for(int i=head[v],u;i;i=bi[i].nxt)
	{
		u=bi[i].to;
		if(u==fa)continue;
		if(leaf[u])++cnt;
		else s=u;
	}
	if(!s)
	{
		for(int i=1;i<=cnt-1;++i)cur[p+i]=p+i+1;
		cur[p+cnt]=p+1;
		cur[p+cnt+1]=p+cnt+1;
	}
	else
	{
		for(int i=1;i<=cnt;++i)cur[p+i]=p+i+1;
		cur[p+cnt+1]=p+1;
		p+=cnt+1;
		dfs(s,v);
	}
}
void check(int v)
{
//	printf("%d\n",v);
	p=0;dfs(v,0);
//	for(int i=1;i<=n;++i)printf("%d ",cur[i]);printf("\n");
	bool ok;
	if(!ans[1])ok=1;
	else
	{
		for(int i=1;i<=n;++i)
			if(cur[i]!=ans[i])
			{
				ok=cur[i]<ans[i];
				break;
			}
	}
	if(ok)memcpy(ans,cur,sizeof(cur));
}
int main()
{
	scanf("%d",&n);
	if(n==2)return printf("1 2"),0;
	int x,y;
	for(int i=1;i<n;++i)
	{
		scanf("%d %d",&x,&y);
		add(x,y),add(y,x);++d[x],++d[y];
	}
	for(int i=1;i<=n;++i)if(d[i]==1)leaf[i]=1;
	for(int i=1;i<=n;++i)if(leaf[i])--d[bi[head[i]].to];
	for(int i=1;i<=n;++i)
		if(!leaf[i]&&d[i]>2)return printf("-1"),0;
	for(int i=1;i<=n;++i)
		if(!leaf[i]&&d[i]<=1)
		{
			check(i);
			for(int j=head[i];j;j=bi[j].nxt)
				if(leaf[bi[j].to]){check(bi[j].to);break;}
		}
	for(int i=1;i<=n;++i)printf("%d ",ans[i]);
	return 0;
}

```