#include<bits/stdc++.h>
using namespace std;
const int N=1e5+5;
int n,st=1,id,len,res[N],ans[N],dis[N],en=1,tot,head[N],nex[N<<1],to[N<<1];
bool vis[N],path[N];
void add(int u,int v){to[++tot]=v;nex[tot]=head[u];head[u]=tot;}
void dfs1(int u,int p)
{
    dis[u]=dis[p]+1;
    if(dis[u]>dis[st]) st=u;
    for(int i=head[u];i;i=nex[i])
    {
        int v=to[i];if(v==p)continue;
        dfs1(v,u);
    }
}
void dfs2(int u,int p)
{
    dis[u]=dis[p]+1;
    if(dis[u]>dis[en]) en=u;
    for(int i=head[u];i;i=nex[i])
    {
        int v=to[i];if(v==p) continue;
        dfs2(v,u);
    }
}
void dfs3(int u,int p)
{
    vis[u]=true;
    if(u==en)
    {
        for(int i=1;i<=n;i++)
            path[i]=vis[i];
    }
    for(int i=head[u];i;i=nex[i])
    {
        int v=to[i];if(v==p) continue;
        dfs3(v,u);
    }
    vis[u]=false;
}
void dfs4(int u,int p)
{
    int s=++id;
    for(int i=head[u];i;i=nex[i])
    {
        int v=to[i];if(v==p) continue;
        if(!path[v])
            ans[++len]=++id;
    }
    ans[++len]=s;
    for(int i=head[u];i;i=nex[i])
    {
        int v=to[i];if(v==p) continue;
        if(path[v])
            dfs4(v,u);
    }
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<n;i++)
    {
        int u,v;scanf("%d%d",&u,&v);
        add(u,v);add(v,u);
    }
    dfs1(st,0);
    dfs2(st,0);
    dfs3(st,0);
    dfs4(st,0);
    if(len!=n){printf("-1\n");return 0;}
    for(int i=1;i<=n;i++)
        res[i]=ans[i];
    len=id=0;
    dfs4(en,0);
    bool flag;
    for(int i=1;i<=n;i++)
        if(res[i]!=ans[i])
        flag=res[i]<ans[i];
    if(flag)
        for(int i=1;i<=n;i++)
        printf(i==n?"%d\n":"%d ",res[i]);
    else for(int i=1;i<=n;i++)
        printf(i==n?"%d\n":"%d ",ans[i]);
}

### Prompt

Generate a Cpp solution 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>
using namespace std;
const int N=1e5+5;
int n,st=1,id,len,res[N],ans[N],dis[N],en=1,tot,head[N],nex[N<<1],to[N<<1];
bool vis[N],path[N];
void add(int u,int v){to[++tot]=v;nex[tot]=head[u];head[u]=tot;}
void dfs1(int u,int p)
{
    dis[u]=dis[p]+1;
    if(dis[u]>dis[st]) st=u;
    for(int i=head[u];i;i=nex[i])
    {
        int v=to[i];if(v==p)continue;
        dfs1(v,u);
    }
}
void dfs2(int u,int p)
{
    dis[u]=dis[p]+1;
    if(dis[u]>dis[en]) en=u;
    for(int i=head[u];i;i=nex[i])
    {
        int v=to[i];if(v==p) continue;
        dfs2(v,u);
    }
}
void dfs3(int u,int p)
{
    vis[u]=true;
    if(u==en)
    {
        for(int i=1;i<=n;i++)
            path[i]=vis[i];
    }
    for(int i=head[u];i;i=nex[i])
    {
        int v=to[i];if(v==p) continue;
        dfs3(v,u);
    }
    vis[u]=false;
}
void dfs4(int u,int p)
{
    int s=++id;
    for(int i=head[u];i;i=nex[i])
    {
        int v=to[i];if(v==p) continue;
        if(!path[v])
            ans[++len]=++id;
    }
    ans[++len]=s;
    for(int i=head[u];i;i=nex[i])
    {
        int v=to[i];if(v==p) continue;
        if(path[v])
            dfs4(v,u);
    }
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<n;i++)
    {
        int u,v;scanf("%d%d",&u,&v);
        add(u,v);add(v,u);
    }
    dfs1(st,0);
    dfs2(st,0);
    dfs3(st,0);
    dfs4(st,0);
    if(len!=n){printf("-1\n");return 0;}
    for(int i=1;i<=n;i++)
        res[i]=ans[i];
    len=id=0;
    dfs4(en,0);
    bool flag;
    for(int i=1;i<=n;i++)
        if(res[i]!=ans[i])
        flag=res[i]<ans[i];
    if(flag)
        for(int i=1;i<=n;i++)
        printf(i==n?"%d\n":"%d ",res[i]);
    else for(int i=1;i<=n;i++)
        printf(i==n?"%d\n":"%d ",ans[i]);
}

```