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

const int N=1e5+5;

struct Edge { int v; Edge *nxt; } edge[N<<1],*head[N];
int n,fa[N],Size[N],per[N],f1[N],f2[N],cnt_edge,cnt_per,max_dist;

inline void addedge(int u,int v){
    edge[++cnt_edge].v=v; Size[u]++;
    edge[cnt_edge].nxt=head[u],head[u]=&edge[cnt_edge];
}

inline void Diam_DFS(int now,int from,int dist,int &max_dist,int &t){
    fa[now]=from;
    if (dist>max_dist) max_dist=dist,t=now;
    for (register Edge *u=head[now];u;u=u->nxt)
        if (u->v!=from) Diam_DFS(u->v,now,dist+1,max_dist,t);
}

inline pair <int,int> Diameter(){
    int s=1,t=1;
    Diam_DFS(s,0,0,max_dist=0,t);
    s=t;
    Diam_DFS(s,0,0,max_dist=0,t);
    return make_pair(s,t);
}

int main(){
    scanf("%d",&n);
    for (register int i=1,u,v;i<n;i++)
        scanf("%d%d",&u,&v),addedge(u,v),addedge(v,u);
    pair <int,int> p=Diameter();
    int s=p.second,t=p.first,past=0;
    while (1){
        cnt_per++;
        for (register Edge *u=head[s];u;u=u->nxt)
            if (Size[u->v]>1&&u->v!=fa[s]&&u->v!=past){ printf("-1\n"); return 0; }
            else if (u->v!=fa[s]&&u->v!=past) per[cnt_per]++;
        if (s==t) break;
        past=s,s=fa[s];
    }
    for (register int i=1,sum1=1,sum2=1;i<=cnt_per;i++,sum1++,sum2++){
        for (register int j=1;j<=per[i];j++,sum1++)
            f1[sum1]=sum1+1; f1[sum1]=sum1-per[i];
        for (register int j=1;j<=per[cnt_per-i+1];j++,sum2++)
            f2[sum2]=sum2+1; f2[sum2]=sum2-per[cnt_per-i+1];
    }
    for (register int i=1;i<=n;i++)
        if (f1[i]<f2[i]){ for (register int j=1;j<=n;j++) printf("%d ",f1[j]); putchar('\n'); return 0; }
        else if (f2[i]<f1[i]){ for (register int j=1;j<=n;j++) printf("%d ",f2[j]); putchar('\n'); return 0; }
    for (register int j=1;j<=n;j++) printf("%d ",f1[j]); putchar('\n');
    return 0;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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;

const int N=1e5+5;

struct Edge { int v; Edge *nxt; } edge[N<<1],*head[N];
int n,fa[N],Size[N],per[N],f1[N],f2[N],cnt_edge,cnt_per,max_dist;

inline void addedge(int u,int v){
    edge[++cnt_edge].v=v; Size[u]++;
    edge[cnt_edge].nxt=head[u],head[u]=&edge[cnt_edge];
}

inline void Diam_DFS(int now,int from,int dist,int &max_dist,int &t){
    fa[now]=from;
    if (dist>max_dist) max_dist=dist,t=now;
    for (register Edge *u=head[now];u;u=u->nxt)
        if (u->v!=from) Diam_DFS(u->v,now,dist+1,max_dist,t);
}

inline pair <int,int> Diameter(){
    int s=1,t=1;
    Diam_DFS(s,0,0,max_dist=0,t);
    s=t;
    Diam_DFS(s,0,0,max_dist=0,t);
    return make_pair(s,t);
}

int main(){
    scanf("%d",&n);
    for (register int i=1,u,v;i<n;i++)
        scanf("%d%d",&u,&v),addedge(u,v),addedge(v,u);
    pair <int,int> p=Diameter();
    int s=p.second,t=p.first,past=0;
    while (1){
        cnt_per++;
        for (register Edge *u=head[s];u;u=u->nxt)
            if (Size[u->v]>1&&u->v!=fa[s]&&u->v!=past){ printf("-1\n"); return 0; }
            else if (u->v!=fa[s]&&u->v!=past) per[cnt_per]++;
        if (s==t) break;
        past=s,s=fa[s];
    }
    for (register int i=1,sum1=1,sum2=1;i<=cnt_per;i++,sum1++,sum2++){
        for (register int j=1;j<=per[i];j++,sum1++)
            f1[sum1]=sum1+1; f1[sum1]=sum1-per[i];
        for (register int j=1;j<=per[cnt_per-i+1];j++,sum2++)
            f2[sum2]=sum2+1; f2[sum2]=sum2-per[cnt_per-i+1];
    }
    for (register int i=1;i<=n;i++)
        if (f1[i]<f2[i]){ for (register int j=1;j<=n;j++) printf("%d ",f1[j]); putchar('\n'); return 0; }
        else if (f2[i]<f1[i]){ for (register int j=1;j<=n;j++) printf("%d ",f2[j]); putchar('\n'); return 0; }
    for (register int j=1;j<=n;j++) printf("%d ",f1[j]); putchar('\n');
    return 0;
}
```