#include <bits/stdc++.h>
using namespace std;
const int MAXN=4e5+7;
int head[MAXN],vis[MAXN],size[MAXN],dep[MAXN],nod[MAXN];
struct po
{
	int nxt,to,from;
}edge[MAXN];
int n,m,num,cnt,ans=1,s,t,flag=1;
inline void add_edge(int from,int to)
{
	edge[++num].nxt=head[from];
	edge[num].from=from;
	edge[num].to=to;
	head[from]=num;
}
inline int bfs(int S)
{
	memset(vis,0,sizeof(vis));
	queue<int> q;
	q.push(S);
	vis[S]=1;
	int u;
	while(!q.empty()){
		u=q.front(); q.pop();
		for(int i=head[u];i;i=edge[i].nxt){
			int v=edge[i].to;
			if(!vis[v]){
				vis[v]=1;
				q.push(v);
			}
		}
	}
	return u;
}
void dfs(int u)
{
	if(!flag) return;
	int tp=0;
	vis[u]=1;
	for(int i=head[u];i;i=edge[i].nxt){
		int v=edge[i].to;
		if(!vis[v]){
			dfs(v);
			dep[u]=max(dep[u],dep[v]);
			if(dep[v]>1) tp++;
			if(v!=t&&dep[v]==1) size[u]++;
		}
	}
	dep[u]++;
	if(tp>=2){
		flag=0;
		return;
	}
	if(u!=s&&u!=t&&dep[u]>=2)
		nod[cnt++]=size[u];
}
void out(int u)
{
	int tt=ans;
	vis[u]=1;
	if(size[u]>0) for(int i=1;i<=size[u];i++) cout<<++ans<<" ";
	printf("%d%c", tt,u==t?'\n':' ');
	ans++;
	for(int i=head[u];i;i=edge[i].nxt){
		int v=edge[i].to;
		if(v==t||v==s||dep[v]>1){
			if(!vis[v]) out(v);
		}
	}
}
int main()
{
	cin>>n;
	m=n-1;
	for(int i=1;i<=m;i++){
		int u,v;
		cin>>u>>v;
		add_edge(u,v);
		add_edge(v,u);
	}
	t=bfs(1);s=bfs(t);
	memset(vis,0,sizeof(vis));
	dfs(s);
	
	reverse(nod,nod+cnt);
	for(int i=0;i<cnt;i++){
		if(nod[i]<nod[cnt-i-1]) break;
		else if(nod[i]>nod[cnt-i-1]){
			swap(s,t);
			break;
		}
	}
	memset(vis,0,sizeof(vis));
	if(!flag) cout<<"-1";
	else out(s);
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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 <bits/stdc++.h>
using namespace std;
const int MAXN=4e5+7;
int head[MAXN],vis[MAXN],size[MAXN],dep[MAXN],nod[MAXN];
struct po
{
	int nxt,to,from;
}edge[MAXN];
int n,m,num,cnt,ans=1,s,t,flag=1;
inline void add_edge(int from,int to)
{
	edge[++num].nxt=head[from];
	edge[num].from=from;
	edge[num].to=to;
	head[from]=num;
}
inline int bfs(int S)
{
	memset(vis,0,sizeof(vis));
	queue<int> q;
	q.push(S);
	vis[S]=1;
	int u;
	while(!q.empty()){
		u=q.front(); q.pop();
		for(int i=head[u];i;i=edge[i].nxt){
			int v=edge[i].to;
			if(!vis[v]){
				vis[v]=1;
				q.push(v);
			}
		}
	}
	return u;
}
void dfs(int u)
{
	if(!flag) return;
	int tp=0;
	vis[u]=1;
	for(int i=head[u];i;i=edge[i].nxt){
		int v=edge[i].to;
		if(!vis[v]){
			dfs(v);
			dep[u]=max(dep[u],dep[v]);
			if(dep[v]>1) tp++;
			if(v!=t&&dep[v]==1) size[u]++;
		}
	}
	dep[u]++;
	if(tp>=2){
		flag=0;
		return;
	}
	if(u!=s&&u!=t&&dep[u]>=2)
		nod[cnt++]=size[u];
}
void out(int u)
{
	int tt=ans;
	vis[u]=1;
	if(size[u]>0) for(int i=1;i<=size[u];i++) cout<<++ans<<" ";
	printf("%d%c", tt,u==t?'\n':' ');
	ans++;
	for(int i=head[u];i;i=edge[i].nxt){
		int v=edge[i].to;
		if(v==t||v==s||dep[v]>1){
			if(!vis[v]) out(v);
		}
	}
}
int main()
{
	cin>>n;
	m=n-1;
	for(int i=1;i<=m;i++){
		int u,v;
		cin>>u>>v;
		add_edge(u,v);
		add_edge(v,u);
	}
	t=bfs(1);s=bfs(t);
	memset(vis,0,sizeof(vis));
	dfs(s);
	
	reverse(nod,nod+cnt);
	for(int i=0;i<cnt;i++){
		if(nod[i]<nod[cnt-i-1]) break;
		else if(nod[i]>nod[cnt-i-1]){
			swap(s,t);
			break;
		}
	}
	memset(vis,0,sizeof(vis));
	if(!flag) cout<<"-1";
	else out(s);
}
```