#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define Fast_IO ios::sync_with_stdio(false);
#define DEBUG fprintf(stderr,"Running on Line %d in Function %s\n",__LINE__,__FUNCTION__)
//mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
#define fir first
#define sec second
#define mod 998244353
#define ll long long
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3f
inline int read()
{
	char ch=getchar(); int nega=1; while(!isdigit(ch)) {if(ch=='-') nega=-1; ch=getchar();}
	int ans=0; while(isdigit(ch)) {ans=ans*10+ch-48;ch=getchar();}
	if(nega==-1) return -ans;
	return ans;
}
typedef pair<int,int> pii;
#define N 100005
vector<int> G[N];
int n,dep[N],fa[N];
void dfs(int u,int f)
{
	dep[u]=dep[f]+1; fa[u]=f;
	for(int v:G[u]) if(v!=f) dfs(v,u);
}
bool ch[N];
vector<int> solve(int u,int v)
{
	dfs(u,0); vector<int> ans; int cnt=0;
	while(v!=u)
	{
		int _cnt=0;
		for(int _:G[v])
		{
			if(!ch[_]) _cnt++;
		}
		int r=++cnt;
		while(_cnt--) ans.pb(++cnt);
		ans.pb(r);
		v=fa[v];
	}
	ans.pb(n);
	return ans;
}
bool cmp(vector<int> x,vector<int> y)
{
	for(int i=0;i<(int)x.size();i++)
	{
		if(x[i]!=y[i]) return x[i]<y[i];
	}
	return 0;
}
signed main()
{
	cin>>n;
	for(int i=1;i<n;i++)
	{
		int u=read(),v=read();
		G[u].pb(v),G[v].pb(u);
	}
	dfs(1,0);
	int maxn=0;
	for(int i=1;i<=n;i++) if(dep[i]>dep[maxn]) maxn=i;
	dfs(maxn,0);
	int amaxn=0;
	for(int i=1;i<=n;i++) if(dep[i]>dep[amaxn]) amaxn=i;
	int r=amaxn;
	while(r!=maxn)
	{
		ch[r]=1;
		r=fa[r];
	}
	ch[maxn]=1;
	bool ok=1;
	for(int i=1;i<=n;i++)
	{
		if(ch[i]) continue;
		if(G[i].size()>1) ok=0;
	}
	if(!ok) {cout<<"-1\n"; return 0;}
	auto x=solve(maxn,amaxn),y=solve(amaxn,maxn);
	if(!cmp(x,y)) swap(x,y);
	for(int i:x) printf("%d%c",i," \n"[i==n]);
	return 0;
}



### 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>
using namespace std;
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define Fast_IO ios::sync_with_stdio(false);
#define DEBUG fprintf(stderr,"Running on Line %d in Function %s\n",__LINE__,__FUNCTION__)
//mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
#define fir first
#define sec second
#define mod 998244353
#define ll long long
#define inf 0x3f3f3f3f
#define INF 0x3f3f3f3f3f3f3f3f
inline int read()
{
	char ch=getchar(); int nega=1; while(!isdigit(ch)) {if(ch=='-') nega=-1; ch=getchar();}
	int ans=0; while(isdigit(ch)) {ans=ans*10+ch-48;ch=getchar();}
	if(nega==-1) return -ans;
	return ans;
}
typedef pair<int,int> pii;
#define N 100005
vector<int> G[N];
int n,dep[N],fa[N];
void dfs(int u,int f)
{
	dep[u]=dep[f]+1; fa[u]=f;
	for(int v:G[u]) if(v!=f) dfs(v,u);
}
bool ch[N];
vector<int> solve(int u,int v)
{
	dfs(u,0); vector<int> ans; int cnt=0;
	while(v!=u)
	{
		int _cnt=0;
		for(int _:G[v])
		{
			if(!ch[_]) _cnt++;
		}
		int r=++cnt;
		while(_cnt--) ans.pb(++cnt);
		ans.pb(r);
		v=fa[v];
	}
	ans.pb(n);
	return ans;
}
bool cmp(vector<int> x,vector<int> y)
{
	for(int i=0;i<(int)x.size();i++)
	{
		if(x[i]!=y[i]) return x[i]<y[i];
	}
	return 0;
}
signed main()
{
	cin>>n;
	for(int i=1;i<n;i++)
	{
		int u=read(),v=read();
		G[u].pb(v),G[v].pb(u);
	}
	dfs(1,0);
	int maxn=0;
	for(int i=1;i<=n;i++) if(dep[i]>dep[maxn]) maxn=i;
	dfs(maxn,0);
	int amaxn=0;
	for(int i=1;i<=n;i++) if(dep[i]>dep[amaxn]) amaxn=i;
	int r=amaxn;
	while(r!=maxn)
	{
		ch[r]=1;
		r=fa[r];
	}
	ch[maxn]=1;
	bool ok=1;
	for(int i=1;i<=n;i++)
	{
		if(ch[i]) continue;
		if(G[i].size()>1) ok=0;
	}
	if(!ok) {cout<<"-1\n"; return 0;}
	auto x=solve(maxn,amaxn),y=solve(amaxn,maxn);
	if(!cmp(x,y)) swap(x,y);
	for(int i:x) printf("%d%c",i," \n"[i==n]);
	return 0;
}



```