#include <bits/stdc++.h>
using namespace std;

vector<int> adjList[100010];
bool leaf[100010];
bool ipt[100010];

vector<int> ans;
vector<int> ans2;
int cnt = 1;
bool possib = true;

vector<int> iadjList[100010];
void build(int u,int p){
	for(auto v: adjList[u]) if(ipt[v]){
		iadjList[u].push_back(v);
		if(v != p) build(v,u);
	}
}

void build2(int u,int p){
	int curr = cnt,nxt = -1;
	cnt += 1;
	for(auto v: adjList[u]) if(v != p){
		if(ipt[v]){
			if(nxt == -1) nxt = v;
			else possib = false;
		}else{
			ans.push_back(cnt++);
		}
	}
	ans.push_back(curr);
	if(nxt == -1) return;
	build2(nxt,u);
}
int main(){
	int N;cin>>N;
	for(int i=0;i<N-1;i++){
		int u,v;cin>>u>>v;u--;v--;
		adjList[u].push_back(v);
		adjList[v].push_back(u);
	}
	
	for(int i=0;i<N;i++) if(adjList[i].size() == 1) leaf[i] = true;
	for(int i=0;i<N;i++) if(leaf[i] == false) ipt[i] = true;
	for(int i=0;i<N;i++) if(ipt[i]){
		build(i,-1);
		break;
	}
	for(int i=0;i<N;i++) if(ipt[i] && (iadjList[i].size() == 1 || iadjList[i].size() == 0)){
		int nxt = (iadjList[i].size() > 0?iadjList[i][0]:-1);
		for(auto v: adjList[i]) if(v != nxt){
			ipt[v] = true;
			iadjList[v].push_back(i);
			iadjList[i].push_back(v);
			if(nxt == -1){
				nxt = v;
			}else break;
		}
	}
	
	for(int i=0;i<N;i++) if(ipt[i] && (iadjList[i].size() == 1 || iadjList[i].size() == 0)){
		if(ans.size() > 0){
			swap(ans,ans2);
			cnt = 1;
		}
		build2(i,-1);
	}

	if(ans2.size() == N && ans2 < ans) swap(ans,ans2);
	if(!possib || ans.size() != N) cout<<-1<<endl;
	else {
		for(auto x: ans) cout<<x<<" ";
		cout<<endl;
	}
}
### Prompt

Develop a solution in CPP 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 <bits/stdc++.h>
using namespace std;

vector<int> adjList[100010];
bool leaf[100010];
bool ipt[100010];

vector<int> ans;
vector<int> ans2;
int cnt = 1;
bool possib = true;

vector<int> iadjList[100010];
void build(int u,int p){
	for(auto v: adjList[u]) if(ipt[v]){
		iadjList[u].push_back(v);
		if(v != p) build(v,u);
	}
}

void build2(int u,int p){
	int curr = cnt,nxt = -1;
	cnt += 1;
	for(auto v: adjList[u]) if(v != p){
		if(ipt[v]){
			if(nxt == -1) nxt = v;
			else possib = false;
		}else{
			ans.push_back(cnt++);
		}
	}
	ans.push_back(curr);
	if(nxt == -1) return;
	build2(nxt,u);
}
int main(){
	int N;cin>>N;
	for(int i=0;i<N-1;i++){
		int u,v;cin>>u>>v;u--;v--;
		adjList[u].push_back(v);
		adjList[v].push_back(u);
	}
	
	for(int i=0;i<N;i++) if(adjList[i].size() == 1) leaf[i] = true;
	for(int i=0;i<N;i++) if(leaf[i] == false) ipt[i] = true;
	for(int i=0;i<N;i++) if(ipt[i]){
		build(i,-1);
		break;
	}
	for(int i=0;i<N;i++) if(ipt[i] && (iadjList[i].size() == 1 || iadjList[i].size() == 0)){
		int nxt = (iadjList[i].size() > 0?iadjList[i][0]:-1);
		for(auto v: adjList[i]) if(v != nxt){
			ipt[v] = true;
			iadjList[v].push_back(i);
			iadjList[i].push_back(v);
			if(nxt == -1){
				nxt = v;
			}else break;
		}
	}
	
	for(int i=0;i<N;i++) if(ipt[i] && (iadjList[i].size() == 1 || iadjList[i].size() == 0)){
		if(ans.size() > 0){
			swap(ans,ans2);
			cnt = 1;
		}
		build2(i,-1);
	}

	if(ans2.size() == N && ans2 < ans) swap(ans,ans2);
	if(!possib || ans.size() != N) cout<<-1<<endl;
	else {
		for(auto x: ans) cout<<x<<" ";
		cout<<endl;
	}
}
```