#include<bits/stdc++.h>
const int N=100005;
using namespace std;
int n,to[N*2],hd[N*2],lk[N],cnt,
re[N],f[N],sz[N],de[N];
bool vis[N];
inline void add(int u,int v){
	to[++cnt]=v,hd[cnt]=lk[u],lk[u]=cnt;
}int u,v;
void dfs(int x){
	sz[x]=1;
	for(int s,i=lk[x];i;i=hd[i])
	if(f[x]^(s=to[i]))
	f[s]=x,de[s]=de[x]+1,dfs(s),
	sz[x]+=sz[s];
}
inline void get(int x){
	f[x]=0,dfs(x);
	for(int i=1;i<=n;i++)
	if(de[i]>de[u])u=i;
}
inline void cal(int s){
	for(int k=1;k<s;k++)
	printf("%d ",u+k);
	printf("%d ",u),u+=s;
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<n;i++)
	scanf("%d%d",&u,&v),
	add(u,v),add(v,u);
	get(1),get(u),v=cnt=0;
	for(;vis[u]=1,u;u=f[u])
	re[cnt++]=sz[u]-v,v=sz[u];
	for(int i=1;i<=n;i++)
	if(!vis[f[i]])return puts("-1"),0;
	u=1,v=2;
	for(int i=0,j=cnt-1;i<cnt;i++,j--){
		if(v>1&&re[i]!=re[j])
		v=re[i]<re[j];
		cal(re[v?i:j]);
	}
}
### 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<bits/stdc++.h>
const int N=100005;
using namespace std;
int n,to[N*2],hd[N*2],lk[N],cnt,
re[N],f[N],sz[N],de[N];
bool vis[N];
inline void add(int u,int v){
	to[++cnt]=v,hd[cnt]=lk[u],lk[u]=cnt;
}int u,v;
void dfs(int x){
	sz[x]=1;
	for(int s,i=lk[x];i;i=hd[i])
	if(f[x]^(s=to[i]))
	f[s]=x,de[s]=de[x]+1,dfs(s),
	sz[x]+=sz[s];
}
inline void get(int x){
	f[x]=0,dfs(x);
	for(int i=1;i<=n;i++)
	if(de[i]>de[u])u=i;
}
inline void cal(int s){
	for(int k=1;k<s;k++)
	printf("%d ",u+k);
	printf("%d ",u),u+=s;
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<n;i++)
	scanf("%d%d",&u,&v),
	add(u,v),add(v,u);
	get(1),get(u),v=cnt=0;
	for(;vis[u]=1,u;u=f[u])
	re[cnt++]=sz[u]-v,v=sz[u];
	for(int i=1;i<=n;i++)
	if(!vis[f[i]])return puts("-1"),0;
	u=1,v=2;
	for(int i=0,j=cnt-1;i<cnt;i++,j--){
		if(v>1&&re[i]!=re[j])
		v=re[i]<re[j];
		cal(re[v?i:j]);
	}
}
```