#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int N=200005;
int n,i,j,k,st,ed,head[N],adj[N],nxt[N],w1[N],w2[N],fa[N],d[N];
bool v[N];
void dfs(int x,int dad,int d)
{
	if(d>j)
	{
		i=x;
		j=d;
	}
	for(int y=head[x];y;y=nxt[y])
		if(adj[y]!=dad)
			dfs(adj[y],x,d+1);
}
void Dfs(int x,int dad)
{
	for(int y=head[x];y;y=nxt[y])
		if(adj[y]!=dad)
		{
			fa[adj[y]]=x;
			Dfs(adj[y],x);
		}
}
int main()
{
	scanf("%d",&n);
	for(i=1;i<n;++i)
	{
		scanf("%d%d",&j,&k);
		adj[i*2-1]=k;
		nxt[i*2-1]=head[j];
		head[j]=i*2-1;
		adj[i*2]=j;
		nxt[i*2]=head[k];
		head[k]=i*2;
		++d[j],++d[k];
	}
	i=1,j=0;
	dfs(1,-1,0);
	st=i;
	j=0;
	dfs(i,-1,0);
	ed=i;
	Dfs(st,-1);
	for(i=ed;i;i=fa[i])
		for(j=head[i];j;j=nxt[j])
			v[adj[j]]=true;
	for(i=1;i<=n;++i)
		if(!v[i])
		{
			printf("-1");
			return 0;
		}
	k=0;
	for(i=ed;i;i=fa[i])
	{
		if(i!=st&&i!=ed)
		{
			for(j=1;j<=d[i]-2;++j)
				w1[k+j]=k+j+1;
		}
		w1[k+(i==st||i==ed?1:d[i]-1)]=k+1;
		k+=(i==st||i==ed?1:d[i]-1);
	}
	swap(st,ed);
	for(i=1;i<=n;++i)
		fa[i]=0,v[i]=false;
	Dfs(st,-1);
	for(i=ed;i;i=fa[i])
		for(j=head[i];j;j=nxt[j])
			v[adj[j]]=true;
	for(i=1;i<=n;++i)
		if(!v[i])
		{
			printf("-1");
			return 0;
		}
	k=0;
	for(i=ed;i;i=fa[i])
	{
		if(i!=st&&i!=ed)
		{
			for(j=1;j<=d[i]-2;++j)
				w2[k+j]=k+j+1;
		}
		w2[k+(i==st||i==ed?1:d[i]-1)]=k+1;
		k+=(i==st||i==ed?1:d[i]-1);
	}
	w1[n+1]=0,w2[n+1]=1;
	for(i=1;i<=n+1;++i)
		if(w1[i]!=w2[i])
		{
			if(w1[i]<w2[i])
			{
				for(i=1;i<=n;++i)
					printf("%d ",w1[i]);
			}
			else
			{
				for(i=1;i<=n;++i)
					printf("%d ",w2[i]);
			}
		}
	return 0;
}

### Prompt

Please create a solution in CPP 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<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int N=200005;
int n,i,j,k,st,ed,head[N],adj[N],nxt[N],w1[N],w2[N],fa[N],d[N];
bool v[N];
void dfs(int x,int dad,int d)
{
	if(d>j)
	{
		i=x;
		j=d;
	}
	for(int y=head[x];y;y=nxt[y])
		if(adj[y]!=dad)
			dfs(adj[y],x,d+1);
}
void Dfs(int x,int dad)
{
	for(int y=head[x];y;y=nxt[y])
		if(adj[y]!=dad)
		{
			fa[adj[y]]=x;
			Dfs(adj[y],x);
		}
}
int main()
{
	scanf("%d",&n);
	for(i=1;i<n;++i)
	{
		scanf("%d%d",&j,&k);
		adj[i*2-1]=k;
		nxt[i*2-1]=head[j];
		head[j]=i*2-1;
		adj[i*2]=j;
		nxt[i*2]=head[k];
		head[k]=i*2;
		++d[j],++d[k];
	}
	i=1,j=0;
	dfs(1,-1,0);
	st=i;
	j=0;
	dfs(i,-1,0);
	ed=i;
	Dfs(st,-1);
	for(i=ed;i;i=fa[i])
		for(j=head[i];j;j=nxt[j])
			v[adj[j]]=true;
	for(i=1;i<=n;++i)
		if(!v[i])
		{
			printf("-1");
			return 0;
		}
	k=0;
	for(i=ed;i;i=fa[i])
	{
		if(i!=st&&i!=ed)
		{
			for(j=1;j<=d[i]-2;++j)
				w1[k+j]=k+j+1;
		}
		w1[k+(i==st||i==ed?1:d[i]-1)]=k+1;
		k+=(i==st||i==ed?1:d[i]-1);
	}
	swap(st,ed);
	for(i=1;i<=n;++i)
		fa[i]=0,v[i]=false;
	Dfs(st,-1);
	for(i=ed;i;i=fa[i])
		for(j=head[i];j;j=nxt[j])
			v[adj[j]]=true;
	for(i=1;i<=n;++i)
		if(!v[i])
		{
			printf("-1");
			return 0;
		}
	k=0;
	for(i=ed;i;i=fa[i])
	{
		if(i!=st&&i!=ed)
		{
			for(j=1;j<=d[i]-2;++j)
				w2[k+j]=k+j+1;
		}
		w2[k+(i==st||i==ed?1:d[i]-1)]=k+1;
		k+=(i==st||i==ed?1:d[i]-1);
	}
	w1[n+1]=0,w2[n+1]=1;
	for(i=1;i<=n+1;++i)
		if(w1[i]!=w2[i])
		{
			if(w1[i]<w2[i])
			{
				for(i=1;i<=n;++i)
					printf("%d ",w1[i]);
			}
			else
			{
				for(i=1;i<=n;++i)
					printf("%d ",w2[i]);
			}
		}
	return 0;
}

```