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;

long long int N, a, b;
vector<long long int> edge[100101], c, d;

int main(){
    scanf("%lld", &N);
    for(int i = 1; i < N; ++i){
        scanf("%lld%lld", &a, &b);
        edge[a - 1].push_back(b - 1);
        edge[b - 1].push_back(a - 1);
    }
    b = -1;
    for(int i = 0; i < N; ++i){
        a = 0;
        if(edge[i].size() < 2)continue;
        for(auto j : edge[i]){
            if(edge[j].size() > 1)++a;
        }
        if(a > 2)return 0 & puts("-1");
        if(a == 1)b = i;
    }
    if(b == -1){
        cout << "1 ";
        for(int i = 2; i < N - 1; ++i){
            cout << i + 1 << " ";
        }
        cout << "2 " << N << " ";
        return 0;
    }
    stack<pair<long long int, long long int>> stc;
    stc.push({b, -1});
    c.push_back(0);
    d.push_back(0);
    c.push_back(edge[b].size() - 2);
    d.push_back(edge[b].size() - 2);
    while(!stc.empty()){
        auto i = stc.top();
        stc.pop();
        //cout << i.first << " " << i.second << endl;
        for(auto j : edge[i.first]){
            if(j != i.second && edge[j].size() > 1){
                c.push_back(edge[j].size() - 2);
                d.push_back(edge[j].size() - 2);
                stc.push({j, i.first});
            }
        }
    }
    c.push_back(0);
    d.push_back(0);
    reverse(d.begin(), d.end());
    if(c > d)swap(c, d);
    int i = 0;
    for(auto j : c){
        ++i;
        for(int k = 1; k <= j; ++k){
            cout << i + k << " ";
        }
        cout << i << " ";
        i += j;
    }
    return 0;
}