#include <string>
#include <vector>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <map>
#include <list>
#include <set>
#include <numeric>
#include <queue>
#include <stack>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <cstring>
#include <climits>
#include <cfloat>
#include <ctime>
#include <complex>
#include <cassert>
#include <array>
#include <bitset> 
#include <unordered_map>

using namespace std;
typedef long long LL;
typedef pair<int,int> P;

const int L=1e5+1;
int n;
vector<int> es[L];
P dfs(int r,int d,int p)
{
	P ret=P(d,r);
	for(auto e:es[r]){
		if(e==p){
			continue;
		}
		ret=max(ret,dfs(e,d+1,r));
	}
	return ret;
}
vector<int> ord;
int main() {
	cin >> n;
	for(int i=0;i<n-1;i++){
		int u,v;
		cin >> u >> v;
		es[u].push_back(v);
		es[v].push_back(u);
	}
	P p=dfs(1,0,0);
	P q=dfs(p.second,0,0);
	vector<int> ret;
	vector<int> ret1;

	int s=p.second;
	int t=q.second;
	for(int k=0;k<2;k++){
		int r=s;
		int pre=0;
		while(1){
			int nx=0;
			int v=ret.size()+1;
			for(auto e:es[r]){
				if(e==pre){
					continue;
				}
				if(es[e].size()>1||e==t){
					if(nx){
						cout << -1 << endl;
						return 0;
					}
					nx=e;
				}
				else{
					ret.push_back(ret.size()+2);
				}
			}
			ret.push_back(v);
			if(nx==0){
				break;
			}
			pre=r;
			r=nx;
		}
		ret.swap(ret1);
		swap(s,t);
	}
	if(ret1<ret){
		ret.swap(ret1);
	}
	auto sp="";
	for(auto v:ret){
		cout << sp << v;
		sp=" ";
	}
	cout << endl;

	return 0;
}


### Prompt

Your challenge is to write 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 <string>
#include <vector>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <map>
#include <list>
#include <set>
#include <numeric>
#include <queue>
#include <stack>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <cstring>
#include <climits>
#include <cfloat>
#include <ctime>
#include <complex>
#include <cassert>
#include <array>
#include <bitset> 
#include <unordered_map>

using namespace std;
typedef long long LL;
typedef pair<int,int> P;

const int L=1e5+1;
int n;
vector<int> es[L];
P dfs(int r,int d,int p)
{
	P ret=P(d,r);
	for(auto e:es[r]){
		if(e==p){
			continue;
		}
		ret=max(ret,dfs(e,d+1,r));
	}
	return ret;
}
vector<int> ord;
int main() {
	cin >> n;
	for(int i=0;i<n-1;i++){
		int u,v;
		cin >> u >> v;
		es[u].push_back(v);
		es[v].push_back(u);
	}
	P p=dfs(1,0,0);
	P q=dfs(p.second,0,0);
	vector<int> ret;
	vector<int> ret1;

	int s=p.second;
	int t=q.second;
	for(int k=0;k<2;k++){
		int r=s;
		int pre=0;
		while(1){
			int nx=0;
			int v=ret.size()+1;
			for(auto e:es[r]){
				if(e==pre){
					continue;
				}
				if(es[e].size()>1||e==t){
					if(nx){
						cout << -1 << endl;
						return 0;
					}
					nx=e;
				}
				else{
					ret.push_back(ret.size()+2);
				}
			}
			ret.push_back(v);
			if(nx==0){
				break;
			}
			pre=r;
			r=nx;
		}
		ret.swap(ret1);
		swap(s,t);
	}
	if(ret1<ret){
		ret.swap(ret1);
	}
	auto sp="";
	for(auto v:ret){
		cout << sp << v;
		sp=" ";
	}
	cout << endl;

	return 0;
}


```