#include <bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<int , int> P2;
typedef pair<pair<int , int> , int> P3;
typedef pair<pair<int , int> , pair<int , int> > P4;
#define PB(a) push_back(a)
#define MP(a , b) make_pair((a) , (b))
#define M3P(a , b , c) make_pair(make_pair((a) , (b)) , (c))
#define M4P(a , b , c , d) make_pair(make_pair((a) , (b)) , make_pair((c) , (d)))
#define repp(i,a,b) for(int i = (int)(a) ; i < (int)(b) ; ++i)
#define repm(i,a,b) for(int i = (int)(a) ; i > (int)(b) ; --i)

const int MC = 1e5 + 3;
int n;
vector<int> V[MC];

int main(){
	cin >> n;
	repp(i,1,n){
		int a,b; cin >> a >> b;
		V[a].push_back(b);
		V[b].push_back(a);
	}
	if(n == 2) return cout << "1 2" << endl , 0;
	int w;
	repp(i,1,n+1){
		int c = 0;
		for(auto z : V[i]) if(V[z].size() > 1) ++c;
		if(c > 2) return cout << -1 << endl , 0;
		if(V[i].size() > 1 && c <= 1) w = i;
	}
	vector<int> u;
	int pr = 0;
	while(w){
		u.push_back(V[w].size()-2);
		int nx = 0;
		for(auto z : V[w]) if(z != pr && V[z].size() > 1) nx = z;
		pr = w; w = nx;
	}
	repp(i,0,u.size()) if(u[i] != u[u.size()-1-i]){
		if(u[i] > u[u.size()-1-i]) reverse(u.begin(),u.end());
		break;
	}
	int s = 2;
	cout << 1 << ' ';
	for(auto z : u){
		repp(i,s+1,s+z+1) cout << i << ' ';
		cout << s << ' ';
		s += z+1;
	}
	cout << n << endl;
	return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 <bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef pair<int , int> P2;
typedef pair<pair<int , int> , int> P3;
typedef pair<pair<int , int> , pair<int , int> > P4;
#define PB(a) push_back(a)
#define MP(a , b) make_pair((a) , (b))
#define M3P(a , b , c) make_pair(make_pair((a) , (b)) , (c))
#define M4P(a , b , c , d) make_pair(make_pair((a) , (b)) , make_pair((c) , (d)))
#define repp(i,a,b) for(int i = (int)(a) ; i < (int)(b) ; ++i)
#define repm(i,a,b) for(int i = (int)(a) ; i > (int)(b) ; --i)

const int MC = 1e5 + 3;
int n;
vector<int> V[MC];

int main(){
	cin >> n;
	repp(i,1,n){
		int a,b; cin >> a >> b;
		V[a].push_back(b);
		V[b].push_back(a);
	}
	if(n == 2) return cout << "1 2" << endl , 0;
	int w;
	repp(i,1,n+1){
		int c = 0;
		for(auto z : V[i]) if(V[z].size() > 1) ++c;
		if(c > 2) return cout << -1 << endl , 0;
		if(V[i].size() > 1 && c <= 1) w = i;
	}
	vector<int> u;
	int pr = 0;
	while(w){
		u.push_back(V[w].size()-2);
		int nx = 0;
		for(auto z : V[w]) if(z != pr && V[z].size() > 1) nx = z;
		pr = w; w = nx;
	}
	repp(i,0,u.size()) if(u[i] != u[u.size()-1-i]){
		if(u[i] > u[u.size()-1-i]) reverse(u.begin(),u.end());
		break;
	}
	int s = 2;
	cout << 1 << ' ';
	for(auto z : u){
		repp(i,s+1,s+z+1) cout << i << ' ';
		cout << s << ' ';
		s += z+1;
	}
	cout << n << endl;
	return 0;
}

```