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

typedef long long ll;
typedef pair<int,int> P;

#define fr first
#define sc second

struct Tree{
	int V;
	vector<int> G[100010];
	int dist[3][100010];
	void dfs(int v,int p,int t){
		dist[t][v] = dist[t][p]+1;
		for(int u: G[v]){
			if(u == p)continue;
			dfs(u,v,t);
		}
	}
	int cnt[100010];
	void solve(){
		int x = 1;
		dist[0][0] = -1;
		dfs(x,0,0);
		int y = 1;
		for(int i = 1 ; i <= V ; i ++){
			if(dist[0][i] > dist[0][y])y = i;
		}
		dist[1][0] = -1;
		dfs(y,0,1);
		int z = 1;
		for(int i = 1 ; i <= V ; i ++){
			if(dist[1][i] > dist[1][z])z = i;
		}
		dist[2][0] = -1;
		dfs(z,0,2);
		for(int i = 0 ; i <= dist[1][z] ; i ++){
			cnt[i] = 0;
		}
		
		/*cout << y << " " << z << endl;
		for(int i = 1 ; i <= V ; i ++){
			cout << dist[1][i] << " " << dist[2][i] << endl;
		}*/
		
		for(int i = 1 ; i <= V ; i ++){
			if(dist[1][i]+dist[2][i] == dist[1][z])continue;
			else if(dist[1][i]+dist[2][i] >= dist[1][z]+4){
				puts("-1");
				return;
			}
			else {
				cnt[dist[1][i]-1] ++;
			}
		}
		
		/*for(int i = 0 ; i <= dist[1][z] ; i ++){
			cout << cnt[i] << " ";
		}cout << endl;*/
		
		vector<int> v;
		for(int i = 0 ; i <= dist[1][z] ; i ++){
			v.push_back(cnt[i]);
		}
		vector<int> u = v; reverse(u.begin(),u.end());
		for(int i = 0 ; i < v.size() ; i ++){
			if(v[i] < u[i])break;
			if(v[i] > u[i]){
				v = u;
				break;
			}
		}
		int cnt = 0;
		int hoge = 1;
		int ret[100010];
		for(int i = 0 ; i+1 < v.size() ; i ++){
			ret[cnt+v[i]+1] = hoge++;
			//printf("%d ",cnt+v[i]+1);
			for(int j = 0 ; j < v[i] ; j ++){
				ret[cnt+j+1] = hoge++;
				//printf("%d ",cnt+j+1);
			}
			cnt += v[i]+1;
		}
		ret[V] = V;
		//printf("%d\n",V);
		
		for(int i = 1 ; i <= V ; i ++){
			printf("%d%c",ret[i],(i==V)?'\n':' ');
		}
	}
}tree;

int main(){
	scanf("%d",&tree.V);
	for(int i = 0 ; i < tree.V-1 ; i ++){
		int a,b;
		scanf("%d%d",&a,&b);
		tree.G[a].push_back(b);
		tree.G[b].push_back(a);
	}
	tree.solve();
}

