#include <bits/stdc++.h>

using namespace std;

int N, X, Y, cnt = 0;
vector<int> AdjList[100005];
int dist[100005], onroute[100005];
vector<int> path, perm1, perm2;

void dfs(int x, int p){
	for (auto it : AdjList[x]){
		if (it == p) continue;
		dist[it] = dist[x] + 1;
		dfs(it,x);
	}
}

int route(int x, int t, int p){
	if (x == t) return 1;
	for (auto it : AdjList[x]){
		if (it == p) continue;
		if (route(it,t,x)){
			path.push_back(it);
			return 1;
		} 
	}
	return 0;
}

int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cin >> N;
	for (int i = 1; i < N; ++i){
		cin >> X >> Y;
		AdjList[X].push_back(Y);
		AdjList[Y].push_back(X);
	}
	dist[1] = 0;
	dfs(1,-1);
	int maxi1 = -1e9, maxdex1 = -1;
	for (int i = 1; i <= N; ++i){
		if (dist[i] > maxi1){
			maxi1 = dist[i];
			maxdex1 = i;
		}
		dist[i] = 0;
	}
	dfs(maxdex1,-1);
	int maxi2 = -1e9, maxdex2 = -1;
	for (int i = 1; i <= N; ++i){
		if (dist[i] > maxi2){
			maxi2 = dist[i];
			maxdex2 = i;
		}
	}
	route(maxdex1,maxdex2,-1);
	path.push_back(maxdex1);
	for (auto it : path) onroute[it] = 1;
	for (auto it : path){
		cnt++;
		for (auto ad : AdjList[it]){
			if (!onroute[ad]) cnt++;
		}
	}
	if (cnt != N){
		cout << -1 << '\n';
		return 0;
	}
	int p = 1, last = 1;
	for (int i = 0; i < path.size(); ++i){
		if (i == 0){
			perm1.push_back(1);
		}
		else{
			for (auto it : AdjList[path[i]]){
				if (!onroute[it]){
					p++;
					perm1.push_back(p);
				}
			}
			perm1.push_back(last);
		}
		p++;
		last = p;
	}
	reverse(path.begin(),path.end());
	p = 1, last = 1;
	for (int i = 0; i < path.size(); ++i){
		if (i == 0){
			perm2.push_back(1);
		}
		else{
			for (auto it : AdjList[path[i]]){
				if (!onroute[it]){
					p++;
					perm2.push_back(p);
				}
			}
			perm2.push_back(last);
		}
		p++;
		last = p;
	}
	if (perm1 > perm2) swap(perm1,perm2);
	for (auto it : perm1){
		cout << it << ' ';
	}
	cout << '\n';
}
### Prompt

Generate 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;

int N, X, Y, cnt = 0;
vector<int> AdjList[100005];
int dist[100005], onroute[100005];
vector<int> path, perm1, perm2;

void dfs(int x, int p){
	for (auto it : AdjList[x]){
		if (it == p) continue;
		dist[it] = dist[x] + 1;
		dfs(it,x);
	}
}

int route(int x, int t, int p){
	if (x == t) return 1;
	for (auto it : AdjList[x]){
		if (it == p) continue;
		if (route(it,t,x)){
			path.push_back(it);
			return 1;
		} 
	}
	return 0;
}

int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cin >> N;
	for (int i = 1; i < N; ++i){
		cin >> X >> Y;
		AdjList[X].push_back(Y);
		AdjList[Y].push_back(X);
	}
	dist[1] = 0;
	dfs(1,-1);
	int maxi1 = -1e9, maxdex1 = -1;
	for (int i = 1; i <= N; ++i){
		if (dist[i] > maxi1){
			maxi1 = dist[i];
			maxdex1 = i;
		}
		dist[i] = 0;
	}
	dfs(maxdex1,-1);
	int maxi2 = -1e9, maxdex2 = -1;
	for (int i = 1; i <= N; ++i){
		if (dist[i] > maxi2){
			maxi2 = dist[i];
			maxdex2 = i;
		}
	}
	route(maxdex1,maxdex2,-1);
	path.push_back(maxdex1);
	for (auto it : path) onroute[it] = 1;
	for (auto it : path){
		cnt++;
		for (auto ad : AdjList[it]){
			if (!onroute[ad]) cnt++;
		}
	}
	if (cnt != N){
		cout << -1 << '\n';
		return 0;
	}
	int p = 1, last = 1;
	for (int i = 0; i < path.size(); ++i){
		if (i == 0){
			perm1.push_back(1);
		}
		else{
			for (auto it : AdjList[path[i]]){
				if (!onroute[it]){
					p++;
					perm1.push_back(p);
				}
			}
			perm1.push_back(last);
		}
		p++;
		last = p;
	}
	reverse(path.begin(),path.end());
	p = 1, last = 1;
	for (int i = 0; i < path.size(); ++i){
		if (i == 0){
			perm2.push_back(1);
		}
		else{
			for (auto it : AdjList[path[i]]){
				if (!onroute[it]){
					p++;
					perm2.push_back(p);
				}
			}
			perm2.push_back(last);
		}
		p++;
		last = p;
	}
	if (perm1 > perm2) swap(perm1,perm2);
	for (auto it : perm1){
		cout << it << ' ';
	}
	cout << '\n';
}
```