#include<bits/stdc++.h>
using namespace std;
const int N=1e5+10;
#define sz(a) int(a.size())

int gi() {
	int x=0,o=1;char ch=getchar();
	while(!isdigit(ch)&&ch!='-') ch=getchar();
	if(ch=='-') o=-1,ch=getchar();
	while(isdigit(ch)) x=x*10+ch-'0',ch=getchar();
	return x*o;
}

int n,mx,p,fa[N],st[N],tp=0,cnt[N];
bool vis[N];
vector<int> E[N],ans;

void dfs(int u,int ff,int dis) {
	fa[u]=ff;
	if(dis>mx) mx=dis,p=u;
	for(auto v:E[u]) if(v!=ff) dfs(v,u,dis+1);
}

vector<int> rev(const vector<int> &a) {
	vector<int> ret(n);
	for(int i=0;i<n;i++) ret[a[i]]=i;
	return ret;
}

bool cmp(const vector<int> &a,const vector<int> &b) {
	for(int i=0;i<n;i++) if(a[i]!=b[i]) return a[i]<b[i];
	return 0;
}

void dfs2(int u,int ff,int dep) {
	++cnt[dep];
	for(auto v:E[u]) if(v^ff) dfs2(v,u,dep+1);
}

void work(int rt) {
	memset(cnt,0,sizeof(cnt));
	dfs2(rt,0,0);
	int s=0,fl=1;vector<int> now;
	for(int i=1;fl;i++) {
		if(!cnt[i]) ++cnt[i],fl=0;
		s+=cnt[i];now.push_back(s-1);
		for(int j=cnt[i];j>1;j--) now.push_back(s-j);
	}
	now=rev(now);
	if(ans.empty()||cmp(now,ans)) ans=now;
}

int main() {
	cin>>n;
	for(int i=1,u,v;i<n;i++) u=gi(),v=gi(),E[u].push_back(v),E[v].push_back(u);
	dfs(1,0,0);mx=0;dfs(p,0,0);
	for(int i=p;i;i=fa[i]) vis[st[++tp]=i]=1;
	for(int i=1;i<=tp;i++)
		for(auto j:E[st[i]])
			if(!vis[j]&&sz(E[j])!=1) return cout<<-1,0;
	work(st[1]);work(st[2]);
	work(st[tp-1]);work(st[tp]);
	for(auto j:E[st[2]])
		if(!vis[j]) { work(j);break; }
	for(auto j:E[st[tp-1]])
		if(!vis[j]) { work(j);break; }
	for(auto x:ans) cout<<x+1<<' ';
	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;
const int N=1e5+10;
#define sz(a) int(a.size())

int gi() {
	int x=0,o=1;char ch=getchar();
	while(!isdigit(ch)&&ch!='-') ch=getchar();
	if(ch=='-') o=-1,ch=getchar();
	while(isdigit(ch)) x=x*10+ch-'0',ch=getchar();
	return x*o;
}

int n,mx,p,fa[N],st[N],tp=0,cnt[N];
bool vis[N];
vector<int> E[N],ans;

void dfs(int u,int ff,int dis) {
	fa[u]=ff;
	if(dis>mx) mx=dis,p=u;
	for(auto v:E[u]) if(v!=ff) dfs(v,u,dis+1);
}

vector<int> rev(const vector<int> &a) {
	vector<int> ret(n);
	for(int i=0;i<n;i++) ret[a[i]]=i;
	return ret;
}

bool cmp(const vector<int> &a,const vector<int> &b) {
	for(int i=0;i<n;i++) if(a[i]!=b[i]) return a[i]<b[i];
	return 0;
}

void dfs2(int u,int ff,int dep) {
	++cnt[dep];
	for(auto v:E[u]) if(v^ff) dfs2(v,u,dep+1);
}

void work(int rt) {
	memset(cnt,0,sizeof(cnt));
	dfs2(rt,0,0);
	int s=0,fl=1;vector<int> now;
	for(int i=1;fl;i++) {
		if(!cnt[i]) ++cnt[i],fl=0;
		s+=cnt[i];now.push_back(s-1);
		for(int j=cnt[i];j>1;j--) now.push_back(s-j);
	}
	now=rev(now);
	if(ans.empty()||cmp(now,ans)) ans=now;
}

int main() {
	cin>>n;
	for(int i=1,u,v;i<n;i++) u=gi(),v=gi(),E[u].push_back(v),E[v].push_back(u);
	dfs(1,0,0);mx=0;dfs(p,0,0);
	for(int i=p;i;i=fa[i]) vis[st[++tp]=i]=1;
	for(int i=1;i<=tp;i++)
		for(auto j:E[st[i]])
			if(!vis[j]&&sz(E[j])!=1) return cout<<-1,0;
	work(st[1]);work(st[2]);
	work(st[tp-1]);work(st[tp]);
	for(auto j:E[st[2]])
		if(!vis[j]) { work(j);break; }
	for(auto j:E[st[tp-1]])
		if(!vis[j]) { work(j);break; }
	for(auto x:ans) cout<<x+1<<' ';
	return 0;
}

```