#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<vector>
#include<algorithm>
using namespace std;
vector<int>G[100010];
int dis[100010];
int par[100010];
bool f[100010];
vector<int>dia;
int n;
void dfs(int x,int p)
{
	for(int i=0;i<G[x].size();i++)
	{
		int y=G[x][i];
		if(y==p)continue;
		dis[y]=dis[x]+1;
		par[y]=x;
		dfs(y,x);
	}
}
vector<int>a,b;
int main()
{
	scanf("%d",&n);
	for(int i=1;i<n;i++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		G[x].push_back(y);
		G[y].push_back(x);
	}
	dfs(1,0);
	int u=0,v=0;
	for(int i=1;i<=n;i++)
	{
		if(!u || dis[i]>dis[u])u=i;
	}
	dis[u]=0;
	par[u]=0;
	dfs(u,0);
	for(int i=1;i<=n;i++)
	{
		if(!v || dis[i]>dis[v])v=i;
	}
	while(v)
	{
		f[v]=true;
		dia.push_back(v);
		v=par[v];
	}
	int lt=1;
	for(int i=0;i<dia.size();i++)
	{
		int cur=dia[i];
		int nw=lt+1;
		for(int j=0;j<G[cur].size();j++)
		{
			int nxt=G[cur][j];
			if(f[nxt])continue;
			if(G[nxt].size()>1)
			{
				printf("-1");
				return 0;
			}
			a.push_back(nw++);
		}
		a.push_back(lt);
		lt=nw;
	}
	reverse(dia.begin(),dia.end());
	lt=1;
	for(int i=0;i<dia.size();i++)
	{
		int cur=dia[i];
		int nw=lt+1;
		for(int j=0;j<G[cur].size();j++)
		{
			int nxt=G[cur][j];
			if(f[nxt])continue;
			b.push_back(nw++);
		}
		b.push_back(lt);
		lt=nw;
	}
	for(int i=0;i<n;i++)
	{
		if(a[i]<b[i])break;
		if(a[i]>b[i])
		{
			a=b;break;
		}
	}
	for(int i=0;i<n;i++)printf("%d ",a[i]);
	return 0;
} 
### 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<iostream>
#include<cstdio>
#include<cstdlib>
#include<vector>
#include<algorithm>
using namespace std;
vector<int>G[100010];
int dis[100010];
int par[100010];
bool f[100010];
vector<int>dia;
int n;
void dfs(int x,int p)
{
	for(int i=0;i<G[x].size();i++)
	{
		int y=G[x][i];
		if(y==p)continue;
		dis[y]=dis[x]+1;
		par[y]=x;
		dfs(y,x);
	}
}
vector<int>a,b;
int main()
{
	scanf("%d",&n);
	for(int i=1;i<n;i++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		G[x].push_back(y);
		G[y].push_back(x);
	}
	dfs(1,0);
	int u=0,v=0;
	for(int i=1;i<=n;i++)
	{
		if(!u || dis[i]>dis[u])u=i;
	}
	dis[u]=0;
	par[u]=0;
	dfs(u,0);
	for(int i=1;i<=n;i++)
	{
		if(!v || dis[i]>dis[v])v=i;
	}
	while(v)
	{
		f[v]=true;
		dia.push_back(v);
		v=par[v];
	}
	int lt=1;
	for(int i=0;i<dia.size();i++)
	{
		int cur=dia[i];
		int nw=lt+1;
		for(int j=0;j<G[cur].size();j++)
		{
			int nxt=G[cur][j];
			if(f[nxt])continue;
			if(G[nxt].size()>1)
			{
				printf("-1");
				return 0;
			}
			a.push_back(nw++);
		}
		a.push_back(lt);
		lt=nw;
	}
	reverse(dia.begin(),dia.end());
	lt=1;
	for(int i=0;i<dia.size();i++)
	{
		int cur=dia[i];
		int nw=lt+1;
		for(int j=0;j<G[cur].size();j++)
		{
			int nxt=G[cur][j];
			if(f[nxt])continue;
			b.push_back(nw++);
		}
		b.push_back(lt);
		lt=nw;
	}
	for(int i=0;i<n;i++)
	{
		if(a[i]<b[i])break;
		if(a[i]>b[i])
		{
			a=b;break;
		}
	}
	for(int i=0;i<n;i++)printf("%d ",a[i]);
	return 0;
} 
```