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;


int n;
vector<int>G[101000], G2[101000], ans, vec;

void dfs(int x,int fff){
    vec.push_back(G[x].size()-1);
    for (auto y:G2[x])
        if (y!=fff)
            dfs(y,x);
}
vector<int> fuck(int s){
    vec.clear();
    vec.push_back(1);
    dfs(s,0);
    vec.push_back(1);
    vector<int>res(n,0);
    int las=0;
    for (auto o:vec){
        int p=las;
        res[las+o-1]=p++;
        for (int i=0;i<o-1;++i)
            res[las+i]=p++;
        las=p;
    }
    return res;
}

int main(){
    cin>>n;
    for (int i=1;i<n;++i){
        int u, v; cin>>u>>v;
        G[u].push_back(v);
        G[v].push_back(u);
    }
    if (n==2){
        puts("1 2");
        return 0;
    }
    for (int i=1;i<=n;++i)
        for (auto j:G[i])
            if (G[i].size()>1&&G[j].size()>1)
                G2[j].push_back(i);
    for (int i=1;i<=n;++i)
        if (G2[i].size()>2){
            puts("-1");
            return 0;
        }
    ans.push_back(n+1);
    for (int i=1;i<=n;++i)
        if (G2[i].size()==1)
            ans=min(ans,fuck(i));
    if (ans[0]==n+1)
        for (int i=1;i<=n;++i)
            if (G[i].size()>1)
                ans=min(ans,fuck(i));
    for (auto o:ans)
        printf("%d ",o+1);
    puts("");
}