#include <bits/stdc++.h>
#define mp make_pair
#define pb push_back
#define pii pair<int,int>
#define LL long long
#define st first
#define nd second
#define endl '\n'
using namespace std;

const int MAXN=100005;
int n,a,b,dep[MAXN],mx,r2,sub[MAXN];
bool ok=1;
vector<int> v[MAXN];

int check(int x,int y,int z) {

	int add;
	sub[x]=1;
	dep[x]=dep[y]+1;
	//cout<<x<<" "<<dep[x]<<" "<<mx<<" "<<dep[mx]<<" "<<y<<" "<<dep[y]<<endl;
	if(z&&dep[x]>dep[mx])
		mx=x;
	int cnt=0;
	for(int i=0;i<v[x].size();++i) {
		if(v[x][i]==y)
			continue;
		add=check(v[x][i],x,z);
		if(add>1) ++cnt;
		sub[x]+=add;
	}
	if(cnt>1) ok=0;
	return sub[x];

}

void solve(int x,int y,int cr,vector<int> &vx) {

	int cur=cr;
	bool ok=0;
	for(int i=0;i<v[x].size();++i) {
		if(v[x][i]==y) continue;
		if(sub[v[x][i]]==1)
			vx.pb(++cur);
		else ok=1;
	}
	if(!ok&&v[x].size()>1) {
		vx.pop_back();
		vx.pb(cr);
		vx.pb(cur);
		return;
	}
	vx.pb(cr);
	for(int i=0;i<v[x].size();++i){
		if(v[x][i]!=y&&sub[v[x][i]]>1)
			solve(v[x][i],x,cur+1,vx);
	}

}

int main() {

	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cin>>n;
	for(int i=0;i<n-1;++i) {
		cin>>a>>b;
		v[a].pb(b);
		v[b].pb(a);
	}
	check(1,0,1);
	r2=mx;
	ok=1;
	mx=0;
	check(r2,0,1);
	if(!ok) {
		cout<<-1<<endl;
		return 0;
	}
	//cout<<r2<<" "<<mx<<endl;
	vector<int> v1,v2;
	check(r2,0,0);
	solve(r2,0,1,v1);
	check(mx,0,0);
	solve(mx,0,1,v2);
	v1=min(v1,v2);
	for(int i=0;i<n;++i) {
		cout<<v1[i]<<" ";
	}
	cout<<endl;

}
### Prompt

Your challenge is to write 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 mp make_pair
#define pb push_back
#define pii pair<int,int>
#define LL long long
#define st first
#define nd second
#define endl '\n'
using namespace std;

const int MAXN=100005;
int n,a,b,dep[MAXN],mx,r2,sub[MAXN];
bool ok=1;
vector<int> v[MAXN];

int check(int x,int y,int z) {

	int add;
	sub[x]=1;
	dep[x]=dep[y]+1;
	//cout<<x<<" "<<dep[x]<<" "<<mx<<" "<<dep[mx]<<" "<<y<<" "<<dep[y]<<endl;
	if(z&&dep[x]>dep[mx])
		mx=x;
	int cnt=0;
	for(int i=0;i<v[x].size();++i) {
		if(v[x][i]==y)
			continue;
		add=check(v[x][i],x,z);
		if(add>1) ++cnt;
		sub[x]+=add;
	}
	if(cnt>1) ok=0;
	return sub[x];

}

void solve(int x,int y,int cr,vector<int> &vx) {

	int cur=cr;
	bool ok=0;
	for(int i=0;i<v[x].size();++i) {
		if(v[x][i]==y) continue;
		if(sub[v[x][i]]==1)
			vx.pb(++cur);
		else ok=1;
	}
	if(!ok&&v[x].size()>1) {
		vx.pop_back();
		vx.pb(cr);
		vx.pb(cur);
		return;
	}
	vx.pb(cr);
	for(int i=0;i<v[x].size();++i){
		if(v[x][i]!=y&&sub[v[x][i]]>1)
			solve(v[x][i],x,cur+1,vx);
	}

}

int main() {

	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cin>>n;
	for(int i=0;i<n-1;++i) {
		cin>>a>>b;
		v[a].pb(b);
		v[b].pb(a);
	}
	check(1,0,1);
	r2=mx;
	ok=1;
	mx=0;
	check(r2,0,1);
	if(!ok) {
		cout<<-1<<endl;
		return 0;
	}
	//cout<<r2<<" "<<mx<<endl;
	vector<int> v1,v2;
	check(r2,0,0);
	solve(r2,0,1,v1);
	check(mx,0,0);
	solve(mx,0,1,v2);
	v1=min(v1,v2);
	for(int i=0;i<n;++i) {
		cout<<v1[i]<<" ";
	}
	cout<<endl;

}
```