#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<vector>
using namespace std;

const int maxn=1e5+10;
int n,ord[maxn],cnt,deg[maxn],ans1[maxn],ans2[maxn];
vector<int> vec[maxn],g[maxn];

int main(){
	scanf("%d",&n);
	for(int i=1,u,v;i<n;++i){
		scanf("%d%d",&u,&v);
		vec[u].push_back(v);
		vec[v].push_back(u);
	}
	if(n<=3){
		for(int i=1;i<=n;++i)
			printf("%d\n",i);
		return 0;
	}
	for(int i=1;i<=n;++i)
		if(vec[i].size()==1)
			++deg[vec[i][0]];
		else
			for(int j=0;j<vec[i].size();++j)
				if(vec[vec[i][j]].size()!=1)
					g[i].push_back(vec[i][j]);
	for(int i=1;i<=n;++i)
		if(g[i].size()>2){
			puts("-1");
			return 0;
		}
	for(int i=1;i<=n;++i)
		if(g[i].size()==1){
			for(int p=i,last=-1;;){
				ord[++cnt]=p;
				bool flag=false;
				for(int j=0;j<g[p].size();++j)
					if(g[p][j]!=last){
						flag=true;
						last=p;
						p=g[p][j];
						break;
					}
				if(!flag)
					break;
			}
			goto fxckglt;
		}
	printf("1 ");
	for(int i=2;i<=n-2;++i)
		printf("%d ",i+1);
	printf("2 %d",n);
	return 0;
fxckglt:
	--deg[ord[1]];
	--deg[ord[cnt]];
	++cnt;
	for(int i=0,top=0;i<=cnt;++i){
		for(int j=0;j<deg[ord[i]];++j)
			++top,ans1[top]=top+1;
		++top;
		ans1[top]=top-deg[ord[i]];
	}
	for(int i=cnt,top=0;~i;--i){
		for(int j=0;j<deg[ord[i]];++j)
			++top,ans2[top]=top+1;
		++top;
		ans2[top]=top-deg[ord[i]];
	}
	for(int i=1;i<=n;++i)
		if(ans1[i]<ans2[i]){
			for(int j=1;j<=n;++j)
				printf("%d ",ans1[j]);
			return 0;
		}
		else if(ans1[i]>ans2[i]){
			for(int j=1;j<=n;++j)
				printf("%d ",ans2[j]);
			return 0;
		}
	for(int i=1;i<=n;++i)
		printf("%d ",ans1[i]);
	return 0;
}
### Prompt

In cpp, your task is to solve 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<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<vector>
using namespace std;

const int maxn=1e5+10;
int n,ord[maxn],cnt,deg[maxn],ans1[maxn],ans2[maxn];
vector<int> vec[maxn],g[maxn];

int main(){
	scanf("%d",&n);
	for(int i=1,u,v;i<n;++i){
		scanf("%d%d",&u,&v);
		vec[u].push_back(v);
		vec[v].push_back(u);
	}
	if(n<=3){
		for(int i=1;i<=n;++i)
			printf("%d\n",i);
		return 0;
	}
	for(int i=1;i<=n;++i)
		if(vec[i].size()==1)
			++deg[vec[i][0]];
		else
			for(int j=0;j<vec[i].size();++j)
				if(vec[vec[i][j]].size()!=1)
					g[i].push_back(vec[i][j]);
	for(int i=1;i<=n;++i)
		if(g[i].size()>2){
			puts("-1");
			return 0;
		}
	for(int i=1;i<=n;++i)
		if(g[i].size()==1){
			for(int p=i,last=-1;;){
				ord[++cnt]=p;
				bool flag=false;
				for(int j=0;j<g[p].size();++j)
					if(g[p][j]!=last){
						flag=true;
						last=p;
						p=g[p][j];
						break;
					}
				if(!flag)
					break;
			}
			goto fxckglt;
		}
	printf("1 ");
	for(int i=2;i<=n-2;++i)
		printf("%d ",i+1);
	printf("2 %d",n);
	return 0;
fxckglt:
	--deg[ord[1]];
	--deg[ord[cnt]];
	++cnt;
	for(int i=0,top=0;i<=cnt;++i){
		for(int j=0;j<deg[ord[i]];++j)
			++top,ans1[top]=top+1;
		++top;
		ans1[top]=top-deg[ord[i]];
	}
	for(int i=cnt,top=0;~i;--i){
		for(int j=0;j<deg[ord[i]];++j)
			++top,ans2[top]=top+1;
		++top;
		ans2[top]=top-deg[ord[i]];
	}
	for(int i=1;i<=n;++i)
		if(ans1[i]<ans2[i]){
			for(int j=1;j<=n;++j)
				printf("%d ",ans1[j]);
			return 0;
		}
		else if(ans1[i]>ans2[i]){
			for(int j=1;j<=n;++j)
				printf("%d ",ans2[j]);
			return 0;
		}
	for(int i=1;i<=n;++i)
		printf("%d ",ans1[i]);
	return 0;
}
```