#include <cstdio>
#include <cstdlib>

int n,head[100001],nxt[200001],b[200001],k,d[100001],con[100001][2],root,cnt,num[100001];
int ans[2][100001];
void push(int s,int t){
	nxt[++k]=head[s];
	head[s]=k;
	b[k]=t;
}
void dfs(int x,int f,bool bb){
	int ch=0;
	for(int i=head[x];i;i=nxt[i])
		if(b[i]!=f){
			++ch;
			if((bb&&ch>2)||(!bb&&ch>1)){puts("-1");exit(0);}
			dfs(b[i],x,0);
		}
	if(!ch)root=x;
}
void dfs(int x,int f){
	num[++cnt]=d[x]-2;
	for(int i=head[x];i;i=nxt[i])
		if(b[i]!=f){
			dfs(b[i],x); 
		}
}
int main(){
	scanf("%d",&n);
	if(n==2){
		printf("1 2\n");
		return 0;
	}
	for(int i=1,u,v;i<n;i++){
		scanf("%d%d",&u,&v);
		con[i][0]=u,con[i][1]=v;
		++d[u],++d[v];
	}
	for(int i=1;i<n;i++)
		if(d[con[i][0]]>1&&d[con[i][1]]>1){
			push(con[i][0],con[i][1]);
			push(con[i][1],con[i][0]);
		}
	for(int i=1;i<=n;i++)
		if(d[i]>1){
			dfs(i,0,1);
			break;
		}
	dfs(root,0);
	num[0]=num[++cnt]=0;
	int top=0;
	for(int i=0,last=1;i<=cnt;i++){
		for(int j=last+1;j<=last+num[i];j++)
			ans[0][++top]=j;
		ans[0][++top]=last;
		last+=1+num[i];
	}
	top=0;
	for(int i=cnt,last=1;~i;i--){
		for(int j=last+1;j<=last+num[i];j++)
			ans[1][++top]=j;
		ans[1][++top]=last;
		last+=1+num[i];
	}
	bool cando=0;
	for(int i=1;i<=n;i++)
		if(ans[0][i]!=ans[1][i]){
			if(ans[0][i]<ans[1][i])cando=0;
			else cando=1;
			break;
		}
	for(int i=1;i<=n;i++)printf("%d ",ans[cando][i]);
}
### Prompt

Please create a solution in cpp 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 <cstdio>
#include <cstdlib>

int n,head[100001],nxt[200001],b[200001],k,d[100001],con[100001][2],root,cnt,num[100001];
int ans[2][100001];
void push(int s,int t){
	nxt[++k]=head[s];
	head[s]=k;
	b[k]=t;
}
void dfs(int x,int f,bool bb){
	int ch=0;
	for(int i=head[x];i;i=nxt[i])
		if(b[i]!=f){
			++ch;
			if((bb&&ch>2)||(!bb&&ch>1)){puts("-1");exit(0);}
			dfs(b[i],x,0);
		}
	if(!ch)root=x;
}
void dfs(int x,int f){
	num[++cnt]=d[x]-2;
	for(int i=head[x];i;i=nxt[i])
		if(b[i]!=f){
			dfs(b[i],x); 
		}
}
int main(){
	scanf("%d",&n);
	if(n==2){
		printf("1 2\n");
		return 0;
	}
	for(int i=1,u,v;i<n;i++){
		scanf("%d%d",&u,&v);
		con[i][0]=u,con[i][1]=v;
		++d[u],++d[v];
	}
	for(int i=1;i<n;i++)
		if(d[con[i][0]]>1&&d[con[i][1]]>1){
			push(con[i][0],con[i][1]);
			push(con[i][1],con[i][0]);
		}
	for(int i=1;i<=n;i++)
		if(d[i]>1){
			dfs(i,0,1);
			break;
		}
	dfs(root,0);
	num[0]=num[++cnt]=0;
	int top=0;
	for(int i=0,last=1;i<=cnt;i++){
		for(int j=last+1;j<=last+num[i];j++)
			ans[0][++top]=j;
		ans[0][++top]=last;
		last+=1+num[i];
	}
	top=0;
	for(int i=cnt,last=1;~i;i--){
		for(int j=last+1;j<=last+num[i];j++)
			ans[1][++top]=j;
		ans[1][++top]=last;
		last+=1+num[i];
	}
	bool cando=0;
	for(int i=1;i<=n;i++)
		if(ans[0][i]!=ans[1][i]){
			if(ans[0][i]<ans[1][i])cando=0;
			else cando=1;
			break;
		}
	for(int i=1;i<=n;i++)printf("%d ",ans[cando][i]);
}
```