#include<iostream>
#include<vector>
using namespace std;
int n;
int deg[100001];
int root;
vector<int>adj[100001];
bool vis[100001];
int p[200001],q[200001];
int ptr=1;
int dfsc=0;
void dfs(int id,int* a,bool r){
	for(int i=ptr+1; i<=ptr+adj[id].size()-r-1 ;i++) swap(a[i],a[i-1]);
	ptr+=adj[id].size()-r;
	vis[id]=true;
	for(int i=0; i<adj[id].size() ;i++){
		int cur=adj[id][i];
		if(deg[cur]!=1 && !vis[cur]) dfs(cur,a,1);
	}
}
void solve(int id){
	dfsc++;
	int* tmp;
	if(dfsc==1) tmp=p;
	else tmp=q;
	for(int i=1; i<=n ;i++) vis[i]=false,tmp[i]=i;
	ptr=1;
	for(int i=0; i<adj[id].size() ;i++){
		if(deg[adj[id][i]]==1){
			dfs(adj[id][i],tmp,0);
			return;
		}
	}
	dfs(id,tmp,0);
}
void out(int* a){
	for(int i=1; i<=n ;i++) cout << a[i] << ' ';
	cout << endl;
}
int main(){
	ios::sync_with_stdio(false);
	cin >> n;
	if(n==2){
		cout << "1 2\n";
		return 0;
	}
	for(int i=2; i<=n ;i++){
		int u,v;
		cin >> u >> v;
		adj[u].push_back(v);
		adj[v].push_back(u);
		deg[u]++;deg[v]++;
	}
	for(int i=1; i<=n ;i++){
		int cnt=0;
		if(deg[i]==1) continue;
		for(int j=0; j<adj[i].size() ;j++) cnt+=(deg[adj[i][j]]!=1);
		if(cnt==0){
			cout << 1 << ' ';
			for(int j=3; j<=n-1 ;j++) cout << j << ' ';
			cout << 2;
			if(n!=2) cout << ' ' << n << endl;
			return 0;
		}
		if(cnt==1){
			solve(i);
		}
		if(cnt>=3){
			cout << -1 << endl;
			return 0;
		}
	}
	for(int i=1; i<=n ;i++){
		if(p[i]<q[i]) out(p);
		else if(p[i]>q[i]) out(q);
		else continue;
		return 0;
	}
	out(p);
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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>
using namespace std;
int n;
int deg[100001];
int root;
vector<int>adj[100001];
bool vis[100001];
int p[200001],q[200001];
int ptr=1;
int dfsc=0;
void dfs(int id,int* a,bool r){
	for(int i=ptr+1; i<=ptr+adj[id].size()-r-1 ;i++) swap(a[i],a[i-1]);
	ptr+=adj[id].size()-r;
	vis[id]=true;
	for(int i=0; i<adj[id].size() ;i++){
		int cur=adj[id][i];
		if(deg[cur]!=1 && !vis[cur]) dfs(cur,a,1);
	}
}
void solve(int id){
	dfsc++;
	int* tmp;
	if(dfsc==1) tmp=p;
	else tmp=q;
	for(int i=1; i<=n ;i++) vis[i]=false,tmp[i]=i;
	ptr=1;
	for(int i=0; i<adj[id].size() ;i++){
		if(deg[adj[id][i]]==1){
			dfs(adj[id][i],tmp,0);
			return;
		}
	}
	dfs(id,tmp,0);
}
void out(int* a){
	for(int i=1; i<=n ;i++) cout << a[i] << ' ';
	cout << endl;
}
int main(){
	ios::sync_with_stdio(false);
	cin >> n;
	if(n==2){
		cout << "1 2\n";
		return 0;
	}
	for(int i=2; i<=n ;i++){
		int u,v;
		cin >> u >> v;
		adj[u].push_back(v);
		adj[v].push_back(u);
		deg[u]++;deg[v]++;
	}
	for(int i=1; i<=n ;i++){
		int cnt=0;
		if(deg[i]==1) continue;
		for(int j=0; j<adj[i].size() ;j++) cnt+=(deg[adj[i][j]]!=1);
		if(cnt==0){
			cout << 1 << ' ';
			for(int j=3; j<=n-1 ;j++) cout << j << ' ';
			cout << 2;
			if(n!=2) cout << ' ' << n << endl;
			return 0;
		}
		if(cnt==1){
			solve(i);
		}
		if(cnt>=3){
			cout << -1 << endl;
			return 0;
		}
	}
	for(int i=1; i<=n ;i++){
		if(p[i]<q[i]) out(p);
		else if(p[i]>q[i]) out(q);
		else continue;
		return 0;
	}
	out(p);
}
```