#include <bits/stdc++.h>
//#pragma GCC diagnostic warning "-std=c++11"
using namespace std;
#define ll long long 
#define pb push_back
//#define mp make_pair
#define fr first
#define sc second
#define MAX ((ll)(1e18+100))
#define MX ((ll)(1e6+90))
#define ARRS ((ll)(1e5+100))
#define MOD ((ll)(1e9+7))
#define EP ((double)(1e-9))
#define EPS ((double)(1e-8))
#define pb push_back
#define PI ((double)3.141592653)
#define LG 16

ll n,m;
ll pas[ARRS];
ll f[ARRS];
vector<ll> v[ARRS];
vector<ll> g[ARRS];

int main(){

	cin>>n;
	if(n==2){
		cout<<"1 2";
		return 0;
	}
	for(int i=0; i<n-1; i++){
		ll k,p;
		cin>>k>>p;
		v[k].pb(p);
		v[p].pb(k);
		f[k]++;
		f[p]++;
	}
	for(int i=1; i<=n; i++){
		if(f[i]>1){
			for(auto x:v[i]){
				if(f[x]>1)
					g[i].pb(x);
			}
			if(g[i].size()>2){
				cout<<-1;
				return 0;
			}
		}
	}
	vector<ll> t;
	for(int i=1; i<=n; i++){
		if(g[i].size()==1){
			ll l=-1;
			ll x=i;
			while(1){
				t.pb(f[x]-g[x].size());
				if(l!=-1&&g[x].size()==1)
					break;
				for(auto y:g[x]){
					if(y==l)continue;
					l=x;
					x=y;
					break;
				}
			}
			break;
		}
	}
	if(t.size()==0){
		t.pb(n-1);
	}
	vector<ll> bk=t;
	ll m=t.size();
	if(t[0]){
		t[0]--;
		reverse(t.begin(),t.end());
		t.pb(0);
		reverse(t.begin(),t.end());
	}
	if(t[t.size()-1]){
		t[t.size()-1]--;
		t.pb(0);
	}
	ll c=1;
	vector<ll> pas;
	for(auto x:t){
		ll k=c++;
		while(x--)
			pas.pb(c++);
		pas.pb(k);
	}


	reverse(bk.begin(),bk.end());
	t=bk;
	if(t[0]){
		t[0]--;
		reverse(t.begin(),t.end());
		t.pb(0);
		reverse(t.begin(),t.end());
	}
	if(t[t.size()-1]){
		t[t.size()-1]--;
		t.pb(0);
	}
	c=1;
	vector<ll> pas1;
	for(auto x:t){
		ll k=c++;
		while(x--)
			pas1.pb(c++);
		pas1.pb(k);
	}
	for(int i=0; i<pas.size(); i++){
		if(pas[i]>pas1[i]){
			pas=pas1;
			break;
		}
		else if(pas[i]<pas1[i]){
			break;
		}
	}
	for(auto x:pas){
		cout<<x<<" ";
	}


}

### Prompt

Please formulate 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>
//#pragma GCC diagnostic warning "-std=c++11"
using namespace std;
#define ll long long 
#define pb push_back
//#define mp make_pair
#define fr first
#define sc second
#define MAX ((ll)(1e18+100))
#define MX ((ll)(1e6+90))
#define ARRS ((ll)(1e5+100))
#define MOD ((ll)(1e9+7))
#define EP ((double)(1e-9))
#define EPS ((double)(1e-8))
#define pb push_back
#define PI ((double)3.141592653)
#define LG 16

ll n,m;
ll pas[ARRS];
ll f[ARRS];
vector<ll> v[ARRS];
vector<ll> g[ARRS];

int main(){

	cin>>n;
	if(n==2){
		cout<<"1 2";
		return 0;
	}
	for(int i=0; i<n-1; i++){
		ll k,p;
		cin>>k>>p;
		v[k].pb(p);
		v[p].pb(k);
		f[k]++;
		f[p]++;
	}
	for(int i=1; i<=n; i++){
		if(f[i]>1){
			for(auto x:v[i]){
				if(f[x]>1)
					g[i].pb(x);
			}
			if(g[i].size()>2){
				cout<<-1;
				return 0;
			}
		}
	}
	vector<ll> t;
	for(int i=1; i<=n; i++){
		if(g[i].size()==1){
			ll l=-1;
			ll x=i;
			while(1){
				t.pb(f[x]-g[x].size());
				if(l!=-1&&g[x].size()==1)
					break;
				for(auto y:g[x]){
					if(y==l)continue;
					l=x;
					x=y;
					break;
				}
			}
			break;
		}
	}
	if(t.size()==0){
		t.pb(n-1);
	}
	vector<ll> bk=t;
	ll m=t.size();
	if(t[0]){
		t[0]--;
		reverse(t.begin(),t.end());
		t.pb(0);
		reverse(t.begin(),t.end());
	}
	if(t[t.size()-1]){
		t[t.size()-1]--;
		t.pb(0);
	}
	ll c=1;
	vector<ll> pas;
	for(auto x:t){
		ll k=c++;
		while(x--)
			pas.pb(c++);
		pas.pb(k);
	}


	reverse(bk.begin(),bk.end());
	t=bk;
	if(t[0]){
		t[0]--;
		reverse(t.begin(),t.end());
		t.pb(0);
		reverse(t.begin(),t.end());
	}
	if(t[t.size()-1]){
		t[t.size()-1]--;
		t.pb(0);
	}
	c=1;
	vector<ll> pas1;
	for(auto x:t){
		ll k=c++;
		while(x--)
			pas1.pb(c++);
		pas1.pb(k);
	}
	for(int i=0; i<pas.size(); i++){
		if(pas[i]>pas1[i]){
			pas=pas1;
			break;
		}
		else if(pas[i]<pas1[i]){
			break;
		}
	}
	for(auto x:pas){
		cout<<x<<" ";
	}


}

```