#include <bits/stdc++.h>
#define int long long
#define INF 1e+9
using namespace std;

vector<int> G[100000];

signed main(){
	int n,s = -1,t = -1;
	vector<int> ans[2];
	cin >> n;
	if(n == 2){
		printf("1 2\n");
		return 0;
	}
	for(int i = 0;i < n - 1;i++){
		int a,b;
		cin >> a >> b; a--;b--;
		G[a].push_back(b);
		G[b].push_back(a);
	}
	for(int i = 0;i < n;i++){
		int cnt = 0;
		for(int v : G[i]){
			if(G[v].size() >= 2) cnt++;
		}
		if(cnt <= 1){
			if(G[i].size() >= 2) (s == -1 ? s : t) = i;
		}
		else if(cnt >= 3){
			cout << -1 << endl;
			return 0;
		}
	}
	for(int i = 0;i < 2;i++){
		ans[i].push_back(1);
		int pos = s,cnt = 1,prev = -1;
		while(1){
			int nxt = -1;
			for(int j = 0;j < (int)G[pos].size() - 2;j++) ans[i].push_back(cnt + j + 2);
			for(int v : G[pos]){
				if(v != prev && G[v].size() >= 2) nxt = v;
			}
			ans[i].push_back(cnt + 1);
			cnt += G[pos].size() - 1;
			if(pos == t || nxt == -1) break;
			prev = pos;
			pos = nxt;
		}
		ans[i].push_back(n);
		if(t != -1) swap(s,t);
	}
	for(int i = 0;i < n;i++){
		if(ans[0][i] > ans[1][i]){
			for(int j = 0;j < n;j++) cout << (j ? " " : "") << ans[1][j];
			cout << endl;
			break;
		}else if(ans[0][i] < ans[1][i] || i == n - 1){
			for(int j = 0;j < n;j++) cout << (j ? " " : "") << ans[0][j];
			cout << endl;
			break;
		}
	}
	return 0;
}
### Prompt

Create a solution in Cpp for 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 int long long
#define INF 1e+9
using namespace std;

vector<int> G[100000];

signed main(){
	int n,s = -1,t = -1;
	vector<int> ans[2];
	cin >> n;
	if(n == 2){
		printf("1 2\n");
		return 0;
	}
	for(int i = 0;i < n - 1;i++){
		int a,b;
		cin >> a >> b; a--;b--;
		G[a].push_back(b);
		G[b].push_back(a);
	}
	for(int i = 0;i < n;i++){
		int cnt = 0;
		for(int v : G[i]){
			if(G[v].size() >= 2) cnt++;
		}
		if(cnt <= 1){
			if(G[i].size() >= 2) (s == -1 ? s : t) = i;
		}
		else if(cnt >= 3){
			cout << -1 << endl;
			return 0;
		}
	}
	for(int i = 0;i < 2;i++){
		ans[i].push_back(1);
		int pos = s,cnt = 1,prev = -1;
		while(1){
			int nxt = -1;
			for(int j = 0;j < (int)G[pos].size() - 2;j++) ans[i].push_back(cnt + j + 2);
			for(int v : G[pos]){
				if(v != prev && G[v].size() >= 2) nxt = v;
			}
			ans[i].push_back(cnt + 1);
			cnt += G[pos].size() - 1;
			if(pos == t || nxt == -1) break;
			prev = pos;
			pos = nxt;
		}
		ans[i].push_back(n);
		if(t != -1) swap(s,t);
	}
	for(int i = 0;i < n;i++){
		if(ans[0][i] > ans[1][i]){
			for(int j = 0;j < n;j++) cout << (j ? " " : "") << ans[1][j];
			cout << endl;
			break;
		}else if(ans[0][i] < ans[1][i] || i == n - 1){
			for(int j = 0;j < n;j++) cout << (j ? " " : "") << ans[0][j];
			cout << endl;
			break;
		}
	}
	return 0;
}
```