#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=l;i<=r;++i)
const int N=1e5+5;
vector<int>lk[N];
int fa[N],deep[N];
int s[N];
int np[N],p1[N],p2[N];

void dfs(int x,int fr)
{
	fa[x]=fr;deep[x]=deep[fr]+1;
	for(auto y:lk[x])
	if(y!=fr)dfs(y,x);
}

int main()
{
	int n;
	cin>>n;
	rep(i,1,n)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		lk[x].push_back(y);lk[y].push_back(x);
	}
	dfs(1,0);
	int x=1;
	rep(i,2,n)
	if(deep[i]>deep[x])x=i;
	dfs(x,0);
	int y=1;
	rep(i,2,n)
	if(deep[i]>deep[y])y=i;
	for(int i=y;i;i=fa[i])s[i]=1;
	rep(i,1,n)
	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]);
	int top=0;
	for(auto i:a)
	{
		np[top+1]=top+i;
		rep(j,1,i-1)np[top+1+j]=top+j;
		top+=i;
	}
	rep(i,1,n)p1[np[i]]=i;
	top=0;
	for(auto it=a.end();--it>=a.begin();)
	{
		int i=*it;
		np[top+1]=top+i;
		rep(j,1,i-1)np[top+1+j]=top+j;
		top+=i;
	}
	rep(i,1,n)p2[np[i]]=i;
	int i=1;
	while(p1[i]==p2[i]&&i<=n)++i;
	if(p1[i]<p2[i])
	{
		rep(i,1,n)printf("%d ",p1[i]);
	}
	else
	{
		rep(i,1,n)printf("%d ",p2[i]);
	}
}

### 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<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=l;i<=r;++i)
const int N=1e5+5;
vector<int>lk[N];
int fa[N],deep[N];
int s[N];
int np[N],p1[N],p2[N];

void dfs(int x,int fr)
{
	fa[x]=fr;deep[x]=deep[fr]+1;
	for(auto y:lk[x])
	if(y!=fr)dfs(y,x);
}

int main()
{
	int n;
	cin>>n;
	rep(i,1,n)
	{
		int x,y;
		scanf("%d%d",&x,&y);
		lk[x].push_back(y);lk[y].push_back(x);
	}
	dfs(1,0);
	int x=1;
	rep(i,2,n)
	if(deep[i]>deep[x])x=i;
	dfs(x,0);
	int y=1;
	rep(i,2,n)
	if(deep[i]>deep[y])y=i;
	for(int i=y;i;i=fa[i])s[i]=1;
	rep(i,1,n)
	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]);
	int top=0;
	for(auto i:a)
	{
		np[top+1]=top+i;
		rep(j,1,i-1)np[top+1+j]=top+j;
		top+=i;
	}
	rep(i,1,n)p1[np[i]]=i;
	top=0;
	for(auto it=a.end();--it>=a.begin();)
	{
		int i=*it;
		np[top+1]=top+i;
		rep(j,1,i-1)np[top+1+j]=top+j;
		top+=i;
	}
	rep(i,1,n)p2[np[i]]=i;
	int i=1;
	while(p1[i]==p2[i]&&i<=n)++i;
	if(p1[i]<p2[i])
	{
		rep(i,1,n)printf("%d ",p1[i]);
	}
	else
	{
		rep(i,1,n)printf("%d ",p2[i]);
	}
}

```