#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int n,tot,Next[200010],head[100010],tree[200010],In[100010],a[100010],cnt,num[100010];
bool vis[100010];
void add(int x,int y)
{
	tot++;
	Next[tot]=head[x];
	head[x]=tot;
	tree[tot]=y;
}
void dfs(int u,bool flag)
{
	vis[u]=1;
	a[++cnt]=u;
	int U=0,Son=0;
	for (int i=head[u];i;i=Next[i])
	{
		int v=tree[i];
		if (vis[v]) continue;
		Son=v;
		if (In[v]!=1)
		{
			if (U) { puts("-1");exit(0);}
				else U=v;
		}else vis[v]=1,num[u]++;
	}
	if (flag&&!U) dfs(Son,0),num[u]--;
		else if (U) dfs(U,1);
}
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);
		In[x]++;In[y]++;
	}
	for (int i=1;i<=n;i++)
	if (In[i]==1)
	{
		int v=tree[head[i]],NUM=0;
		for (int j=head[v];j;j=Next[j])
			if (In[tree[j]]>=2) NUM++;
		if (NUM<2) { dfs(i,(n!=1));break;}
	}
	if (!cnt) { puts("-1");return 0;}
	bool flag=0;
	for (int i=1;i<=cnt/2;i++)
	{
		if (num[a[i]]>num[a[cnt-i+1]]) { flag=1;break;}
		if (num[a[i]]<num[a[cnt-i+1]]) break;
	}
	if (flag)
	{
		for (int i=1;i<=cnt/2;i++) swap(a[i],a[cnt-i+1]);
	}
	int now=0;
	for (int i=1;i<=cnt;i++)
	{
		int x=++now;
		for (int j=1;j<=num[a[i]];j++) printf("%d ",++now);
		printf("%d%c",x," \n"[i==cnt]);
	}
	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<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
int n,tot,Next[200010],head[100010],tree[200010],In[100010],a[100010],cnt,num[100010];
bool vis[100010];
void add(int x,int y)
{
	tot++;
	Next[tot]=head[x];
	head[x]=tot;
	tree[tot]=y;
}
void dfs(int u,bool flag)
{
	vis[u]=1;
	a[++cnt]=u;
	int U=0,Son=0;
	for (int i=head[u];i;i=Next[i])
	{
		int v=tree[i];
		if (vis[v]) continue;
		Son=v;
		if (In[v]!=1)
		{
			if (U) { puts("-1");exit(0);}
				else U=v;
		}else vis[v]=1,num[u]++;
	}
	if (flag&&!U) dfs(Son,0),num[u]--;
		else if (U) dfs(U,1);
}
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);
		In[x]++;In[y]++;
	}
	for (int i=1;i<=n;i++)
	if (In[i]==1)
	{
		int v=tree[head[i]],NUM=0;
		for (int j=head[v];j;j=Next[j])
			if (In[tree[j]]>=2) NUM++;
		if (NUM<2) { dfs(i,(n!=1));break;}
	}
	if (!cnt) { puts("-1");return 0;}
	bool flag=0;
	for (int i=1;i<=cnt/2;i++)
	{
		if (num[a[i]]>num[a[cnt-i+1]]) { flag=1;break;}
		if (num[a[i]]<num[a[cnt-i+1]]) break;
	}
	if (flag)
	{
		for (int i=1;i<=cnt/2;i++) swap(a[i],a[cnt-i+1]);
	}
	int now=0;
	for (int i=1;i<=cnt;i++)
	{
		int x=++now;
		for (int j=1;j<=num[a[i]];j++) printf("%d ",++now);
		printf("%d%c",x," \n"[i==cnt]);
	}
	return 0;
}

```