#include<bits/stdc++.h>
#define ll long long
using namespace std;
template<typename tn> void read(tn &a){
	tn x=0,f=1; char c=' ';
	for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
	for(;isdigit(c);c=getchar()) x=x*10+c-'0';
	a=x*f;
}
const int N = 101000;
int n,d[N];
vector<int> e[N],V;
void dfs(int x,int dad){
	V.push_back(x);
	for(int v:e[x]) if(v!=dad&&e[v].size()>1){
		dfs(v,x);return;
	}
}
int main(){
	read(n);
	if(n==2){
		cout<<1<<' '<<2<<'\n';
		return 0;
	}
	for(int i=1;i<n;i++){
		int x,y;read(x);read(y);
		e[x].push_back(y);
		e[y].push_back(x);
	}
	int sum=0;
	for(int i=1;i<=n;i++)
		if(e[i].size()>1&&++sum)
			for(int v:e[i]) d[v]++;
	for(int i=1;i<=n;i++){
		if(e[i].size()>1&&d[i]<=1){
			for(int v:e[i]) if(e[v].size()==1){dfs(v,0);break;}
			break;
		}
	}
	if(sum+1!=V.size()){puts("-1");return 0;}
	for(int v:e[V.back()]) if(e[v].size()==1&&v!=V[0]){V.push_back(v);break;}
	for(int i=0;i<V.size();i++)
		if(e[V[i]].size()>e[V[V.size()-i-1]].size()){
			reverse(V.begin(),V.end());
			break;
		}
		else if(e[V[i]].size()<e[V[V.size()-i-1]].size()) break;
	int vis[N]={};
	for(int x:V) vis[x]=1;
	vector<int> ans;
	int now=0;
	for(int x:V){
		int k=++now;
		for(int v:e[x])
			if(!vis[v]){
				ans.push_back(++now);
			}
		ans.push_back(k);
	}
	for(int i=0;i<n;i++)
		cout<<ans[i]<<' ';
	cout<<'\n';
	return 0;
}

### Prompt

Your task is to create 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>
#define ll long long
using namespace std;
template<typename tn> void read(tn &a){
	tn x=0,f=1; char c=' ';
	for(;!isdigit(c);c=getchar()) if(c=='-') f=-1;
	for(;isdigit(c);c=getchar()) x=x*10+c-'0';
	a=x*f;
}
const int N = 101000;
int n,d[N];
vector<int> e[N],V;
void dfs(int x,int dad){
	V.push_back(x);
	for(int v:e[x]) if(v!=dad&&e[v].size()>1){
		dfs(v,x);return;
	}
}
int main(){
	read(n);
	if(n==2){
		cout<<1<<' '<<2<<'\n';
		return 0;
	}
	for(int i=1;i<n;i++){
		int x,y;read(x);read(y);
		e[x].push_back(y);
		e[y].push_back(x);
	}
	int sum=0;
	for(int i=1;i<=n;i++)
		if(e[i].size()>1&&++sum)
			for(int v:e[i]) d[v]++;
	for(int i=1;i<=n;i++){
		if(e[i].size()>1&&d[i]<=1){
			for(int v:e[i]) if(e[v].size()==1){dfs(v,0);break;}
			break;
		}
	}
	if(sum+1!=V.size()){puts("-1");return 0;}
	for(int v:e[V.back()]) if(e[v].size()==1&&v!=V[0]){V.push_back(v);break;}
	for(int i=0;i<V.size();i++)
		if(e[V[i]].size()>e[V[V.size()-i-1]].size()){
			reverse(V.begin(),V.end());
			break;
		}
		else if(e[V[i]].size()<e[V[V.size()-i-1]].size()) break;
	int vis[N]={};
	for(int x:V) vis[x]=1;
	vector<int> ans;
	int now=0;
	for(int x:V){
		int k=++now;
		for(int v:e[x])
			if(!vis[v]){
				ans.push_back(++now);
			}
		ans.push_back(k);
	}
	for(int i=0;i<n;i++)
		cout<<ans[i]<<' ';
	cout<<'\n';
	return 0;
}

```