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
#include <bits/stdc++.h>
using namespace std;
const int Maxn=100005;
int len,sz[Maxn];
vector<int>e[Maxn],path;
vector<int>gen(){
    vector<int>ans;
    for(int i=1,j=0;i<=len;i++){
        int cur=++j;
        for(int k=1;k<=sz[i];k++)ans.push_back(++j);
        ans.push_back(cur);
    }
    return ans;
}
pair<int,int>dfs1(int x,int fa,int dis){
    pair<int,int>ret(dis,x);
    for(int y:e[x]){
        if(y==fa)continue;
        ret=max(ret,dfs1(y,x,dis+1));
    }
    return ret;
}
bool dfs2(int x,int fa,int t){
    if(x==t)return path.push_back(x),true;
    for(int y:e[x]){
        if(y==fa)continue;
        if(dfs2(y,x,t))return path.push_back(x),true;
    }
    return false;
}
int main(){
    int n;scanf("%d",&n);
    for(int i=1;i<n;i++){
        int x,y;scanf("%d%d",&x,&y);
        e[x].push_back(y);
        e[y].push_back(x);
    }
    int a=dfs1(1,0,1).second;
    int b=dfs1(a,0,1).second;
    dfs2(a,0,b);
    len=path.size();
    for(int i=1;i<=len;i++)
        sz[i]=e[path[i-1]].size()-1-(i!=1&&i!=len),sz[0]+=sz[i];
    if(sz[0]+len!=n){
        cout<<"-1\n";
        return 0;
    }
    auto ans1=gen();
    reverse(sz+1,sz+len+1);
    auto ans2=gen();
    auto ans=min(ans1,ans2);
    for(int i=0;i<n;i++)
        cout<<ans[i]<<" \n"[i==n-1];
    return 0;
}