#include<bits/stdc++.h>
using namespace std;
const int N=2e5+5;
int n,mx,rx,ry,cnt,now,top,b[N],v[N],f[N],g[N],fp[N],gp[N],nex[N],head[N];
void addedges(int x,int y){
	nex[++now]=head[x];
	head[x]=now,v[now]=y;
}
void dfs(int x,int fa){
	for(int i=head[x];i;i=nex[i])
		if(v[i]!=fa){
			dfs(v[i],x),++f[v[i]];
			if(f[v[i]]>f[x])	g[x]=f[x],gp[x]=fp[x],f[x]=f[v[i]],fp[x]=fp[v[i]];
			else	if(g[x]<f[v[i]])	g[x]=f[v[i]],gp[x]=fp[v[i]];
		}
	if(f[x]+g[x]>mx)	mx=f[x]+g[x],rx=fp[x],ry=gp[x];
}
int s[N],t[N],son[N],deep[N];
void Dfs(int x,int fa){
	deep[x]=0,son[x]=0;
	for(int i=head[x];i;i=nex[i])
		if(v[i]!=fa){
			Dfs(v[i],x);
			if(deep[son[x]]<deep[v[i]])	son[x]=v[i];
		}
	deep[x]=deep[son[x]]+1;
}
void get_ans(int x,int fa){
	int tmp=++cnt,k=b[x]-(x!=fa);
	if(son[x])	--k;
	while(k--) s[++top]=++cnt;
	s[++top]=tmp;
	for(int i=head[x];i;i=nex[i])
		if(v[i]!=fa&&v[i]!=son[x]&&b[v[i]]!=1)
			puts("-1"),exit(0);
	if(son[x])	get_ans(son[x],x);
}
bool check(){
	for(int i=1;i<=n;i++)
		if(s[i]>t[i])	return false;
		else	if(s[i]<t[i])	return true;
	return false;
}
void pri(int rt){
	cnt=top=0,Dfs(rt,rt),get_ans(rt,rt);
	if(check())	for(int i=1;i<=n;i++)	t[i]=s[i];
}
int main(){
	scanf("%d",&n);
	for(int i=1,x,y;i<n;i++){
		scanf("%d%d",&x,&y);
		addedges(x,y),++b[x];
		addedges(y,x),++b[y];
	}
	for(int i=1;i<=n;i++)	fp[i]=gp[i]=i,t[i]=n;
	dfs(1,1),pri(rx),pri(ry);
	for(int i=1;i<=n;i++)	printf("%d ",t[i]);
	cout<<endl;
	return 0;
}

### Prompt

Generate 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<bits/stdc++.h>
using namespace std;
const int N=2e5+5;
int n,mx,rx,ry,cnt,now,top,b[N],v[N],f[N],g[N],fp[N],gp[N],nex[N],head[N];
void addedges(int x,int y){
	nex[++now]=head[x];
	head[x]=now,v[now]=y;
}
void dfs(int x,int fa){
	for(int i=head[x];i;i=nex[i])
		if(v[i]!=fa){
			dfs(v[i],x),++f[v[i]];
			if(f[v[i]]>f[x])	g[x]=f[x],gp[x]=fp[x],f[x]=f[v[i]],fp[x]=fp[v[i]];
			else	if(g[x]<f[v[i]])	g[x]=f[v[i]],gp[x]=fp[v[i]];
		}
	if(f[x]+g[x]>mx)	mx=f[x]+g[x],rx=fp[x],ry=gp[x];
}
int s[N],t[N],son[N],deep[N];
void Dfs(int x,int fa){
	deep[x]=0,son[x]=0;
	for(int i=head[x];i;i=nex[i])
		if(v[i]!=fa){
			Dfs(v[i],x);
			if(deep[son[x]]<deep[v[i]])	son[x]=v[i];
		}
	deep[x]=deep[son[x]]+1;
}
void get_ans(int x,int fa){
	int tmp=++cnt,k=b[x]-(x!=fa);
	if(son[x])	--k;
	while(k--) s[++top]=++cnt;
	s[++top]=tmp;
	for(int i=head[x];i;i=nex[i])
		if(v[i]!=fa&&v[i]!=son[x]&&b[v[i]]!=1)
			puts("-1"),exit(0);
	if(son[x])	get_ans(son[x],x);
}
bool check(){
	for(int i=1;i<=n;i++)
		if(s[i]>t[i])	return false;
		else	if(s[i]<t[i])	return true;
	return false;
}
void pri(int rt){
	cnt=top=0,Dfs(rt,rt),get_ans(rt,rt);
	if(check())	for(int i=1;i<=n;i++)	t[i]=s[i];
}
int main(){
	scanf("%d",&n);
	for(int i=1,x,y;i<n;i++){
		scanf("%d%d",&x,&y);
		addedges(x,y),++b[x];
		addedges(y,x),++b[y];
	}
	for(int i=1;i<=n;i++)	fp[i]=gp[i]=i,t[i]=n;
	dfs(1,1),pri(rx),pri(ry);
	for(int i=1;i<=n;i++)	printf("%d ",t[i]);
	cout<<endl;
	return 0;
}

```