#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define maxn 200005
using namespace std;
int n,r,res=-1,tot,len,flag;
int head[maxn],nex[maxn],to[maxn];
int q[maxn],pa[maxn],dep[maxn],vis[maxn];
void add(int x,int y)
{
	to[++tot]=y; nex[tot]=head[x]; head[x]=tot;
}
void dfs(int now,int fa)
{
	if(dep[now]>res)
	{
		res=dep[now];
		r=now;
	}

	for(int i=head[now];i;i=nex[i])
	{
		if(to[i]!=fa)
		{
			if(flag)
			pa[to[i]]=now;
			dep[to[i]]=dep[now]+1;
			dfs(to[i],now);
		}
	}
}
int ans1[maxn],ans2[maxn];
int solve(int *a)
{
	int sz=0;
	for(int i=1;i<=tot;i++)
	{
		int now=q[i],temp=0;
		for(int j=head[now];j;j=nex[j])
			if(!vis[to[j]])
				temp++;
		int s=sz;
		for(int j=1;j<=temp;j++)
		{
			a[++sz]=s+j+1;
		}
		a[++sz]=s+1;
	}
	return sz;
}
int judge()
{
	for(int i=1;i<=n;i++)
	{
		if(ans1[i]<ans2[i])
			return true;
		if(ans1[i]>ans2[i])
			return false;
	}
	return true;
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<n;i++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		add(x,y); add(y,x);
	}

	dfs(1,-1); res=-1; dep[r]=0; flag=1;
	dfs(r,-1);

//	cerr<<r<<endl;

	tot=0;

	while(r!=0)
	{
		q[++tot]=r;
		vis[r]=1;
		r=pa[r];
	}

	if(solve(ans1)<n)
	{
		printf("-1\n");
		return 0;
	}

	for(int i=1;i<=tot/2;i++)
		swap(q[i],q[tot-i+1]);

	solve(ans2);

	if(judge())
	{
		for(int i=1;i<=n;i++)
			printf("%d ",ans1[i]);
	}
	else
	{
		for(int i=1;i<=n;i++)
			printf("%d ",ans2[i]);
	}
	printf("\n");
}

### 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<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define maxn 200005
using namespace std;
int n,r,res=-1,tot,len,flag;
int head[maxn],nex[maxn],to[maxn];
int q[maxn],pa[maxn],dep[maxn],vis[maxn];
void add(int x,int y)
{
	to[++tot]=y; nex[tot]=head[x]; head[x]=tot;
}
void dfs(int now,int fa)
{
	if(dep[now]>res)
	{
		res=dep[now];
		r=now;
	}

	for(int i=head[now];i;i=nex[i])
	{
		if(to[i]!=fa)
		{
			if(flag)
			pa[to[i]]=now;
			dep[to[i]]=dep[now]+1;
			dfs(to[i],now);
		}
	}
}
int ans1[maxn],ans2[maxn];
int solve(int *a)
{
	int sz=0;
	for(int i=1;i<=tot;i++)
	{
		int now=q[i],temp=0;
		for(int j=head[now];j;j=nex[j])
			if(!vis[to[j]])
				temp++;
		int s=sz;
		for(int j=1;j<=temp;j++)
		{
			a[++sz]=s+j+1;
		}
		a[++sz]=s+1;
	}
	return sz;
}
int judge()
{
	for(int i=1;i<=n;i++)
	{
		if(ans1[i]<ans2[i])
			return true;
		if(ans1[i]>ans2[i])
			return false;
	}
	return true;
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<n;i++)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		add(x,y); add(y,x);
	}

	dfs(1,-1); res=-1; dep[r]=0; flag=1;
	dfs(r,-1);

//	cerr<<r<<endl;

	tot=0;

	while(r!=0)
	{
		q[++tot]=r;
		vis[r]=1;
		r=pa[r];
	}

	if(solve(ans1)<n)
	{
		printf("-1\n");
		return 0;
	}

	for(int i=1;i<=tot/2;i++)
		swap(q[i],q[tot-i+1]);

	solve(ans2);

	if(judge())
	{
		for(int i=1;i<=n;i++)
			printf("%d ",ans1[i]);
	}
	else
	{
		for(int i=1;i<=n;i++)
			printf("%d ",ans2[i]);
	}
	printf("\n");
}

```