#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
struct Edge{
	int v,nxt;
}e[200010];int tot;
int first[100010],deg[100010];
void build(int u,int v)
{
	e[++tot]=(Edge){v,first[u]};
	first[u]=tot;deg[u]++;
	return ;
}
int p[100010];
int p2[100010];
int dis[100010];
int S;
void dfs(int u,int lst)
{
	dis[u]=dis[lst]+1;
	if(dis[u]>dis[S])S=u;
	for(int i=first[u];i;i=e[i].nxt)
		if(e[i].v!=lst)
			dfs(e[i].v,u);
	return ;
}
int a[100010],num;
bool book[100010];
bool dfs2(int u,int lst,int tar)
{
	if(u==tar)
	{
		book[u]=true;
		return true;
	}
	for(int i=first[u];i;i=e[i].nxt)
		if(e[i].v!=lst){
			bool flag=dfs2(e[i].v,u,tar);
			if(flag){
				a[++num]=u;
				book[u]=true;
				return true;
			}
		}
	return false;
}
int main()
{
	int n;scanf("%d",&n);
	for(int i=1;i<n;i++)
	{
		int u,v;
		scanf("%d %d",&u,&v);
		build(u,v);
		build(v,u);
	}
	S=0;
	dfs(1,0);int s=S;S=0;
	dfs(s,0);int t=S;
	dfs2(s,0,t);
	for(int i=1;i<=n;i++)
	{
		if(book[i])continue;
		bool flag=false;
		for(int j=first[i];j;j=e[j].nxt)
			if(book[e[j].v])flag=true;
		if(!flag){
			printf("-1\n");
			return 0;
		}
	}
	int now=0;
	p[++now]=n;int cur=n;
	for(int i=1;i<num;i++)
	{
		int t=deg[a[i]]-2;
		p[++now]=cur-t-1;
		while(t--)
			p[++now]=--cur;
		cur--;
	}
	p[++now]=1;
	reverse(p+1,p+n+1);
	now=0;
	p2[++now]=n;cur=n;
	for(int i=num-1;i>=1;i--)
	{
		int t=deg[a[i]]-2;
		p2[++now]=cur-t-1;
		while(t--)
			p2[++now]=--cur;
		cur--;
	}
	p2[++now]=1;
	reverse(p2+1,p2+n+1);
	for(int i=1;i<=n;i++)
		if(p[i]>p2[i]){
			swap(p,p2);
			break;
		}
		else if(p[i]<p2[i])
			break;
	for(int i=1;i<=n;i++)
		printf("%d ",p[i]);
	return 0;
}
### Prompt

Create a solution in CPP for 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<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
struct Edge{
	int v,nxt;
}e[200010];int tot;
int first[100010],deg[100010];
void build(int u,int v)
{
	e[++tot]=(Edge){v,first[u]};
	first[u]=tot;deg[u]++;
	return ;
}
int p[100010];
int p2[100010];
int dis[100010];
int S;
void dfs(int u,int lst)
{
	dis[u]=dis[lst]+1;
	if(dis[u]>dis[S])S=u;
	for(int i=first[u];i;i=e[i].nxt)
		if(e[i].v!=lst)
			dfs(e[i].v,u);
	return ;
}
int a[100010],num;
bool book[100010];
bool dfs2(int u,int lst,int tar)
{
	if(u==tar)
	{
		book[u]=true;
		return true;
	}
	for(int i=first[u];i;i=e[i].nxt)
		if(e[i].v!=lst){
			bool flag=dfs2(e[i].v,u,tar);
			if(flag){
				a[++num]=u;
				book[u]=true;
				return true;
			}
		}
	return false;
}
int main()
{
	int n;scanf("%d",&n);
	for(int i=1;i<n;i++)
	{
		int u,v;
		scanf("%d %d",&u,&v);
		build(u,v);
		build(v,u);
	}
	S=0;
	dfs(1,0);int s=S;S=0;
	dfs(s,0);int t=S;
	dfs2(s,0,t);
	for(int i=1;i<=n;i++)
	{
		if(book[i])continue;
		bool flag=false;
		for(int j=first[i];j;j=e[j].nxt)
			if(book[e[j].v])flag=true;
		if(!flag){
			printf("-1\n");
			return 0;
		}
	}
	int now=0;
	p[++now]=n;int cur=n;
	for(int i=1;i<num;i++)
	{
		int t=deg[a[i]]-2;
		p[++now]=cur-t-1;
		while(t--)
			p[++now]=--cur;
		cur--;
	}
	p[++now]=1;
	reverse(p+1,p+n+1);
	now=0;
	p2[++now]=n;cur=n;
	for(int i=num-1;i>=1;i--)
	{
		int t=deg[a[i]]-2;
		p2[++now]=cur-t-1;
		while(t--)
			p2[++now]=--cur;
		cur--;
	}
	p2[++now]=1;
	reverse(p2+1,p2+n+1);
	for(int i=1;i<=n;i++)
		if(p[i]>p2[i]){
			swap(p,p2);
			break;
		}
		else if(p[i]<p2[i])
			break;
	for(int i=1;i<=n;i++)
		printf("%d ",p[i]);
	return 0;
}
```