#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 = 100010;
int n;
vector<int> V[MC];

int main(){
	cin >> n;
	repp(i,1,n){
		int a,b; cin >> a >> b;
		V[a].PB(b);
		V[b].PB(a);
	}
	if(n == 2){
		cout << "1 2" << endl;
		return 0;
	}
	if(n == 3){
		cout << "1 2 3" << endl;
		return 0;
	}
	vector<int> r;
	bool b = 0;
	vector<int> c(n+1,0);
	repm(i,n,0){
		int z = 0;
		for(auto u : V[i]){
			if((int)V[u].size() > 1) ++z;
			else ++c[i];
		}
		if(z > 2) b = 1;
		else if(z < 2 && (int)V[i].size() > 1) r.PB(i);
	}
	if(b) return cout << -1 << endl , 0;
	vector<int> ans(n);
	ans[0] = n+1;
	for(auto _r : r){
		vector<int> q(n);
		int x = _r;
		int d = 1;
		--c[_r];
		q[0] = 0;
		vector<int> b(n+1,0);
		while(d < n){
			if(c[x] < 0) break;
			repp(i,0,c[x]){
				q[d+1+i] = d+i;
			}
			q[d] = d+c[x];
			d += c[x]+1;
			b[x] = 1;
			for(auto u : V[x]){
				if(b[u] == 0 && (int)V[u].size() > 1){
					x = u;
					break;
				}
			}
		}
		++c[_r];
		vector<int> p(n);
		repp(i,0,n) p[q[i]] = i+1;
		swap(p[n-2],p[n-1]);
		repp(i,0,n){
			if(ans[i] == p[i]) continue;
			if(ans[i] > p[i]) swap(ans,p);
			break;
		}
	}
	repp(i,0,n) cout << ans[i] << ' ';
	cout << endl;
	return 0;
}

### Prompt

Generate 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 <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 = 100010;
int n;
vector<int> V[MC];

int main(){
	cin >> n;
	repp(i,1,n){
		int a,b; cin >> a >> b;
		V[a].PB(b);
		V[b].PB(a);
	}
	if(n == 2){
		cout << "1 2" << endl;
		return 0;
	}
	if(n == 3){
		cout << "1 2 3" << endl;
		return 0;
	}
	vector<int> r;
	bool b = 0;
	vector<int> c(n+1,0);
	repm(i,n,0){
		int z = 0;
		for(auto u : V[i]){
			if((int)V[u].size() > 1) ++z;
			else ++c[i];
		}
		if(z > 2) b = 1;
		else if(z < 2 && (int)V[i].size() > 1) r.PB(i);
	}
	if(b) return cout << -1 << endl , 0;
	vector<int> ans(n);
	ans[0] = n+1;
	for(auto _r : r){
		vector<int> q(n);
		int x = _r;
		int d = 1;
		--c[_r];
		q[0] = 0;
		vector<int> b(n+1,0);
		while(d < n){
			if(c[x] < 0) break;
			repp(i,0,c[x]){
				q[d+1+i] = d+i;
			}
			q[d] = d+c[x];
			d += c[x]+1;
			b[x] = 1;
			for(auto u : V[x]){
				if(b[u] == 0 && (int)V[u].size() > 1){
					x = u;
					break;
				}
			}
		}
		++c[_r];
		vector<int> p(n);
		repp(i,0,n) p[q[i]] = i+1;
		swap(p[n-2],p[n-1]);
		repp(i,0,n){
			if(ans[i] == p[i]) continue;
			if(ans[i] > p[i]) swap(ans,p);
			break;
		}
	}
	repp(i,0,n) cout << ans[i] << ' ';
	cout << endl;
	return 0;
}

```