#include<cstdio>
#define M 200002
#define N 100001
int a[N],b[M],c[M],d[N],e[N],f[N],i,l,n,q[N],r,u,v,w,s,t;
int main()
{
	scanf("%d",&n);
	for(i=1;i<n;i++)scanf("%d%d",&u,&v),b[++t]=a[u],c[a[u]=t]=v,b[++t]=a[v],c[a[v]=t]=u;
	for(q[l=1]=v=1,r=2;l<r;)for(i=a[u=q[l++]],v=e[v]<e[u]?u:v;i;i=b[i])if(c[i]!=f[u])e[q[r++]=c[i]]=e[f[c[i]]=u]+1;
	for(e[q[l=1]=w=v]=f[v]=0,r=2;l<r;)for(i=a[u=q[l++]],w=e[w]<e[u]?u:w;i;i=b[i])if(c[i]!=f[u])e[q[r++]=c[i]]=e[f[c[i]]=u]+1;
	for(u=w,t=0;u;u=f[u])for(i=a[u],t++;i;i=b[i])if(c[i]!=f[u])d[t]++,s++;
	if(d[1]++,++s!=n)return 0*puts("-1");
	for(u=1,v=t,s=0;v&&d[u]==d[v];u++)v--;if(v&&d[u]>d[v])for(u=1,v=t;u<v;u++,v--)d[u]^=d[v]^=d[u]^=d[v];
	for(i=1;i<=t;printf("%d%c",s-d[i]+1,i==t?'\n':' '),i++)for(s+=u=d[i];--u;)printf("%d ",s-u+1);
	return 0;
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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>
#define M 200002
#define N 100001
int a[N],b[M],c[M],d[N],e[N],f[N],i,l,n,q[N],r,u,v,w,s,t;
int main()
{
	scanf("%d",&n);
	for(i=1;i<n;i++)scanf("%d%d",&u,&v),b[++t]=a[u],c[a[u]=t]=v,b[++t]=a[v],c[a[v]=t]=u;
	for(q[l=1]=v=1,r=2;l<r;)for(i=a[u=q[l++]],v=e[v]<e[u]?u:v;i;i=b[i])if(c[i]!=f[u])e[q[r++]=c[i]]=e[f[c[i]]=u]+1;
	for(e[q[l=1]=w=v]=f[v]=0,r=2;l<r;)for(i=a[u=q[l++]],w=e[w]<e[u]?u:w;i;i=b[i])if(c[i]!=f[u])e[q[r++]=c[i]]=e[f[c[i]]=u]+1;
	for(u=w,t=0;u;u=f[u])for(i=a[u],t++;i;i=b[i])if(c[i]!=f[u])d[t]++,s++;
	if(d[1]++,++s!=n)return 0*puts("-1");
	for(u=1,v=t,s=0;v&&d[u]==d[v];u++)v--;if(v&&d[u]>d[v])for(u=1,v=t;u<v;u++,v--)d[u]^=d[v]^=d[u]^=d[v];
	for(i=1;i<=t;printf("%d%c",s-d[i]+1,i==t?'\n':' '),i++)for(s+=u=d[i];--u;)printf("%d ",s-u+1);
	return 0;
}
```