#include<bits/stdc++.h>
using namespace std;
const int N=100005;
int top,fa[N],dep[N],s[N],np[N],p1[N],p2[N];
vector<int>g[N];
void dfs(int x,int p)
{
	fa[x]=p;dep[x]=dep[p]+1;
	for(int i=0;i<g[x].size();i++)if(g[x][i]!=p)dfs(g[x][i],x);
}
int main()
{
	int n;scanf("%d",&n);
	for(int i=1,u,v;i<n;i++){scanf("%d%d",&u,&v);g[u].push_back(v);g[v].push_back(u);}
	dfs(1,0);int x=1;
	for(int i=2;i<=n;i++)if(dep[i]>dep[x])x=i;
	dfs(x,0);int y=1;
	for(int i=2;i<=n;i++)if(dep[i]>dep[y])y=i;
	for(int i=y;i;i=fa[i])s[i]=1;
	for(int i=1;i<=n;i++)if(!s[i]&&(++s[fa[i]])==1){puts("-1");exit(0);}
	vector<int>a;
	for(int i=y;i;i=fa[i])a.push_back(s[i]);
	top=0;
	for(int i=0;i<a.size();i++)
	{
		np[top+1]=top+a[i];
		for(int j=1;j<a[i];j++)np[top+j+1]=top+j;
		top+=a[i];
	}
	for(int i=1;i<=n;i++)p1[np[i]]=i;
	top=0;
	for(int i=a.size()-1;i>=0;i--)
	{
		np[top+1]=top+a[i];
		for(int j=1;j<a[i];j++)np[top+j+1]=top+j;
		top+=a[i];
	}
	for(int i=1;i<=n;i++)p2[np[i]]=i;
	int i=1;
	while(p1[i]==p2[i]&&i<=n)i++;
	if(p1[i]<p2[i])for(int i=1;i<=n;i++)printf("%d ",p1[i]);
	else for(int i=1;i<=n;i++)printf("%d ",p2[i]);
	return 0;
}
### Prompt

Create a solution in cpp for 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 N=100005;
int top,fa[N],dep[N],s[N],np[N],p1[N],p2[N];
vector<int>g[N];
void dfs(int x,int p)
{
	fa[x]=p;dep[x]=dep[p]+1;
	for(int i=0;i<g[x].size();i++)if(g[x][i]!=p)dfs(g[x][i],x);
}
int main()
{
	int n;scanf("%d",&n);
	for(int i=1,u,v;i<n;i++){scanf("%d%d",&u,&v);g[u].push_back(v);g[v].push_back(u);}
	dfs(1,0);int x=1;
	for(int i=2;i<=n;i++)if(dep[i]>dep[x])x=i;
	dfs(x,0);int y=1;
	for(int i=2;i<=n;i++)if(dep[i]>dep[y])y=i;
	for(int i=y;i;i=fa[i])s[i]=1;
	for(int i=1;i<=n;i++)if(!s[i]&&(++s[fa[i]])==1){puts("-1");exit(0);}
	vector<int>a;
	for(int i=y;i;i=fa[i])a.push_back(s[i]);
	top=0;
	for(int i=0;i<a.size();i++)
	{
		np[top+1]=top+a[i];
		for(int j=1;j<a[i];j++)np[top+j+1]=top+j;
		top+=a[i];
	}
	for(int i=1;i<=n;i++)p1[np[i]]=i;
	top=0;
	for(int i=a.size()-1;i>=0;i--)
	{
		np[top+1]=top+a[i];
		for(int j=1;j<a[i];j++)np[top+j+1]=top+j;
		top+=a[i];
	}
	for(int i=1;i<=n;i++)p2[np[i]]=i;
	int i=1;
	while(p1[i]==p2[i]&&i<=n)i++;
	if(p1[i]<p2[i])for(int i=1;i<=n;i++)printf("%d ",p1[i]);
	else for(int i=1;i<=n;i++)printf("%d ",p2[i]);
	return 0;
}
```