#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+10;
int c[maxn],du[maxn];
int n;
struct edge{int v,nxt;}e[maxn<<1];
int head[maxn],ecnt=0,to[maxn];
void add(int u,int v)
{
	to[u]=v,to[v]=u;
	e[++ecnt]=(edge){v,head[u]},head[u]=ecnt;
	e[++ecnt]=(edge){u,head[v]},head[v]=ecnt;
}
bool vis[maxn];
vector<int>ans;
int find(int u)
{
	for(int i=head[u];~i;i=e[i].nxt)
	{
		int v=e[i].v;
		if(vis[v]||du[v]==1)continue;
		return v;
	}
}
vector<int>solve(int l,int r)
{
	vector<int>cnt;
	vector<int>ret;
	ret.push_back(1);
	cnt.push_back(du[l]-2);
	memset(vis,0,sizeof(vis));
	while(l!=r)vis[l]=1,l=find(l),cnt.push_back(du[l]-2);
	int id=1;
	for(auto t:cnt)
	{
		for(int i=1;i<=t;i++)ret.push_back(id+i+1);
		ret.push_back(id+1),id+=t+1;
	}
	ret.push_back(n);
	return ret;
	
}
vector<int>min(vector<int>a,vector<int>b)
{
	for(int i=0;i<a.size();i++)
	if(a[i]<b[i])return a;
	else if(a[i]>b[i])return b;
	return a;
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<n;i++)
	{
		int u,v;
		scanf("%d%d",&u,&v);
		add(u,v);
		du[u]++,du[v]++;
	}
	for(int i=1;i<=n;i++)c[i]=du[i];
	for(int i=1;i<=n;i++)if(du[i]==1)c[i]--,c[to[i]]--;
	int mx=0;
	for(int i=1;i<=n;i++)mx=max(mx,c[i]);
	if(mx>2){printf("-1\n");return 0;}
	if(mx==0)
	{
		printf("1 ");
		for(int i=3;i<n;i++)printf("%d ",i);
		printf("2 ");
		printf("%d\n",n);
		return 0;
	}
	int u,v;
	for(int i=1;i<=n;i++)if(c[i]==1)u=i;
	for(int i=n;i>=1;i--)if(c[i]==1)v=i;
	ans=min(solve(u,v),solve(v,u));
	for(auto u:ans)printf("%d ",u);
	printf("\n");
}
### 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<bits/stdc++.h>
using namespace std;
const int maxn=1e5+10;
int c[maxn],du[maxn];
int n;
struct edge{int v,nxt;}e[maxn<<1];
int head[maxn],ecnt=0,to[maxn];
void add(int u,int v)
{
	to[u]=v,to[v]=u;
	e[++ecnt]=(edge){v,head[u]},head[u]=ecnt;
	e[++ecnt]=(edge){u,head[v]},head[v]=ecnt;
}
bool vis[maxn];
vector<int>ans;
int find(int u)
{
	for(int i=head[u];~i;i=e[i].nxt)
	{
		int v=e[i].v;
		if(vis[v]||du[v]==1)continue;
		return v;
	}
}
vector<int>solve(int l,int r)
{
	vector<int>cnt;
	vector<int>ret;
	ret.push_back(1);
	cnt.push_back(du[l]-2);
	memset(vis,0,sizeof(vis));
	while(l!=r)vis[l]=1,l=find(l),cnt.push_back(du[l]-2);
	int id=1;
	for(auto t:cnt)
	{
		for(int i=1;i<=t;i++)ret.push_back(id+i+1);
		ret.push_back(id+1),id+=t+1;
	}
	ret.push_back(n);
	return ret;
	
}
vector<int>min(vector<int>a,vector<int>b)
{
	for(int i=0;i<a.size();i++)
	if(a[i]<b[i])return a;
	else if(a[i]>b[i])return b;
	return a;
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<n;i++)
	{
		int u,v;
		scanf("%d%d",&u,&v);
		add(u,v);
		du[u]++,du[v]++;
	}
	for(int i=1;i<=n;i++)c[i]=du[i];
	for(int i=1;i<=n;i++)if(du[i]==1)c[i]--,c[to[i]]--;
	int mx=0;
	for(int i=1;i<=n;i++)mx=max(mx,c[i]);
	if(mx>2){printf("-1\n");return 0;}
	if(mx==0)
	{
		printf("1 ");
		for(int i=3;i<n;i++)printf("%d ",i);
		printf("2 ");
		printf("%d\n",n);
		return 0;
	}
	int u,v;
	for(int i=1;i<=n;i++)if(c[i]==1)u=i;
	for(int i=n;i>=1;i--)if(c[i]==1)v=i;
	ans=min(solve(u,v),solve(v,u));
	for(auto u:ans)printf("%d ",u);
	printf("\n");
}
```