#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+999;
vector<int>v[maxn];
int n,deg[maxn],dis[maxn],pos1,pos2,mx;
void dfs1(int x,int fa){
	dis[x]=dis[fa]+1;
	if(dis[x]>mx)mx=dis[x],pos1=x;
	for(auto y:v[x]){
		if(y==fa)continue;
		dfs1(y,x);
	}
}
void dfs2(int x,int fa){
	dis[x]=dis[fa]+1;
	if(dis[x]>mx)mx=dis[x],pos2=x;
	for(auto y:v[x]){
		if(y==fa)continue;
		dfs2(y,x);
	}
}
int tot,ans1[maxn],now,ans2[maxn],fl;
void dfs(int x,int fa){
	int cnt=0;
	for(auto y:v[x]){
		if(deg[y]==1&&y!=pos1&&y!=pos2)++cnt;
	}
	if(!fl){
		int t=now+1;++now;
		for(int i=1;i<=cnt;++i)ans1[++tot]=now+1,++now;
		ans1[++tot]=t;
	}else{
		int t=now+1;++now;
		for(int i=1;i<=cnt;++i)ans2[++tot]=now+1,++now;
		ans2[++tot]=t;
	}
	for(auto y:v[x]){
		if(y==fa)continue;
		if(deg[y]>1||y==pos1||y==pos2){
			dfs(y,x);
		}
	}
}
int vis[maxn],stac[maxn],top,flag;
void dfs3(int x){
	stac[++top]=x;vis[x]=1;
	if(x==pos2){flag=1;return;}
	for(auto y:v[x]){
		if(vis[y])continue;
		dfs3(y);
		if(flag)return;
	}
	--top;
}
int main(){
	cin>>n;
	for(int i=1,x,y;i<n;++i){
		cin>>x>>y;v[x].push_back(y);v[y].push_back(x);++deg[x];++deg[y];
	}
	dfs1(1,0);
	mx=0;
	memset(dis,0,sizeof(dis));
	dfs2(pos1,0);
	
	
	dfs3(pos1);
	memset(vis,0,sizeof(vis));
	for(int i=1;i<=top;++i)vis[stac[i]]=1;
	for(int i=1;i<=n;++i){
		if(!vis[i]&&deg[i]>1){
			cout<<-1;exit(0);	
		}
	}
	
	
	fl=0;
	dfs(pos1,0);
	fl=1;tot=0,now=0;
	dfs(pos2,0);
	for(int i=1;i<=n;++i){
		if(ans1[i]!=ans2[i]){
			if(ans1[i]<ans2[i]){
				for(int j=1;j<=n;++j)cout<<ans1[j]<<" ";
				exit(0);
			}else{
				for(int j=1;j<=n;++j)cout<<ans2[j]<<" ";
				exit(0);
			}
		}
	}
	for(int i=1;i<=n;++i)cout<<ans1[i]<<" ";
} 
### Prompt

Develop a solution in CPP 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=1e5+999;
vector<int>v[maxn];
int n,deg[maxn],dis[maxn],pos1,pos2,mx;
void dfs1(int x,int fa){
	dis[x]=dis[fa]+1;
	if(dis[x]>mx)mx=dis[x],pos1=x;
	for(auto y:v[x]){
		if(y==fa)continue;
		dfs1(y,x);
	}
}
void dfs2(int x,int fa){
	dis[x]=dis[fa]+1;
	if(dis[x]>mx)mx=dis[x],pos2=x;
	for(auto y:v[x]){
		if(y==fa)continue;
		dfs2(y,x);
	}
}
int tot,ans1[maxn],now,ans2[maxn],fl;
void dfs(int x,int fa){
	int cnt=0;
	for(auto y:v[x]){
		if(deg[y]==1&&y!=pos1&&y!=pos2)++cnt;
	}
	if(!fl){
		int t=now+1;++now;
		for(int i=1;i<=cnt;++i)ans1[++tot]=now+1,++now;
		ans1[++tot]=t;
	}else{
		int t=now+1;++now;
		for(int i=1;i<=cnt;++i)ans2[++tot]=now+1,++now;
		ans2[++tot]=t;
	}
	for(auto y:v[x]){
		if(y==fa)continue;
		if(deg[y]>1||y==pos1||y==pos2){
			dfs(y,x);
		}
	}
}
int vis[maxn],stac[maxn],top,flag;
void dfs3(int x){
	stac[++top]=x;vis[x]=1;
	if(x==pos2){flag=1;return;}
	for(auto y:v[x]){
		if(vis[y])continue;
		dfs3(y);
		if(flag)return;
	}
	--top;
}
int main(){
	cin>>n;
	for(int i=1,x,y;i<n;++i){
		cin>>x>>y;v[x].push_back(y);v[y].push_back(x);++deg[x];++deg[y];
	}
	dfs1(1,0);
	mx=0;
	memset(dis,0,sizeof(dis));
	dfs2(pos1,0);
	
	
	dfs3(pos1);
	memset(vis,0,sizeof(vis));
	for(int i=1;i<=top;++i)vis[stac[i]]=1;
	for(int i=1;i<=n;++i){
		if(!vis[i]&&deg[i]>1){
			cout<<-1;exit(0);	
		}
	}
	
	
	fl=0;
	dfs(pos1,0);
	fl=1;tot=0,now=0;
	dfs(pos2,0);
	for(int i=1;i<=n;++i){
		if(ans1[i]!=ans2[i]){
			if(ans1[i]<ans2[i]){
				for(int j=1;j<=n;++j)cout<<ans1[j]<<" ";
				exit(0);
			}else{
				for(int j=1;j<=n;++j)cout<<ans2[j]<<" ";
				exit(0);
			}
		}
	}
	for(int i=1;i<=n;++i)cout<<ans1[i]<<" ";
} 
```