#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int N;
int dist[100010] = {};
vector<vector<int>> v(100010);
vector<int> deg;

void dfs(int n,int m){
    for(auto x:v[n]){
        if(x!=m){
            dist[x] = dist[n]+1;
            dfs(x,n);
        }
    }
}

void dfs2(int n,int m){
    deg.push_back((int)v[n].size());
    for(auto x:v[n]){
        if(x!=m && (int)v[x].size()!=1) dfs2(x,n);
    }
}

int main(){
    cin >> N;
    int a,b;
    for(int i=0;i<N-1;i++){
        cin >> a >> b;
        v[a].push_back(b);
        v[b].push_back(a);
    }
    int cnt = 0;
    for(int i=1;i<=N;i++){
        cnt += max((int)v[i].size()-2,0);
    }
    dist[1] = 0; dfs(1,0);
    int ma = 0,id = 1;
    for(int i=1;i<=N;i++){
        if(ma<dist[i]){
            ma = dist[i]; id = i;
        }
    }
    dist[id] = 0; dfs(id,0);
    ma = 0;
    int id2 = -1;
    for(int i=1;i<=N;i++){
        if(ma<dist[i]){
            ma = dist[i]; id2 = i;
        }
    }
    if(N-1-cnt!=ma){
        cout << -1 << endl;
        return 0;
    }
    dfs2(id,0);
    deg.push_back(1);
    int l = 0,r = (int) deg.size()-1;
    while(l<r){
        if(deg[l]!=deg[r]){
            if(deg[l]>deg[r]){
                reverse(deg.begin(),deg.end());
            }
            break;
        }
        l++; r--;
    }
    //deg.pop_back();
    vector<int> ans;
    for(auto x:deg) cerr << x << endl;
    for(int i=0;i<=N;i++) ans.push_back(i);
    int now = 1;
    for(int i=0;i<deg.size();i++){
        if(deg[i]<=2) now++;
        else{
            int n = deg[i]-2;
            reverse(ans.begin()+now,ans.begin()+now+n+1);
            reverse(ans.begin()+now,ans.begin()+now+n);
            now += n+1;
        }
    }
    for(int i=1;i<=N;i++) cout << ans[i] << (i!=N? " ":"\n");
}
### Prompt

Create a solution in cpp for 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 <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int N;
int dist[100010] = {};
vector<vector<int>> v(100010);
vector<int> deg;

void dfs(int n,int m){
    for(auto x:v[n]){
        if(x!=m){
            dist[x] = dist[n]+1;
            dfs(x,n);
        }
    }
}

void dfs2(int n,int m){
    deg.push_back((int)v[n].size());
    for(auto x:v[n]){
        if(x!=m && (int)v[x].size()!=1) dfs2(x,n);
    }
}

int main(){
    cin >> N;
    int a,b;
    for(int i=0;i<N-1;i++){
        cin >> a >> b;
        v[a].push_back(b);
        v[b].push_back(a);
    }
    int cnt = 0;
    for(int i=1;i<=N;i++){
        cnt += max((int)v[i].size()-2,0);
    }
    dist[1] = 0; dfs(1,0);
    int ma = 0,id = 1;
    for(int i=1;i<=N;i++){
        if(ma<dist[i]){
            ma = dist[i]; id = i;
        }
    }
    dist[id] = 0; dfs(id,0);
    ma = 0;
    int id2 = -1;
    for(int i=1;i<=N;i++){
        if(ma<dist[i]){
            ma = dist[i]; id2 = i;
        }
    }
    if(N-1-cnt!=ma){
        cout << -1 << endl;
        return 0;
    }
    dfs2(id,0);
    deg.push_back(1);
    int l = 0,r = (int) deg.size()-1;
    while(l<r){
        if(deg[l]!=deg[r]){
            if(deg[l]>deg[r]){
                reverse(deg.begin(),deg.end());
            }
            break;
        }
        l++; r--;
    }
    //deg.pop_back();
    vector<int> ans;
    for(auto x:deg) cerr << x << endl;
    for(int i=0;i<=N;i++) ans.push_back(i);
    int now = 1;
    for(int i=0;i<deg.size();i++){
        if(deg[i]<=2) now++;
        else{
            int n = deg[i]-2;
            reverse(ans.begin()+now,ans.begin()+now+n+1);
            reverse(ans.begin()+now,ans.begin()+now+n);
            now += n+1;
        }
    }
    for(int i=1;i<=N;i++) cout << ans[i] << (i!=N? " ":"\n");
}
```