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<cstdio>
#include<algorithm>
#include<vector>
using namespace std;
const int N=1e5+5;
int n,fa[N],dep[N],mrk[N],len,t[N];
vector<int>E[N];
void dfs(int u,int f){
	fa[u]=f;dep[u]=dep[f]+1;
	for(int v:E[u])if(v!=f)dfs(v,u);
}
int main(){
	scanf("%d",&n);
	for(int i=1,x,y;i<n;++i){
		scanf("%d%d",&x,&y);
		E[x].push_back(y);E[y].push_back(x);
	}
	dfs(1,0);int P=0;
	for(int i=1;i<=n;++i)P=dep[i]>dep[P]?i:P;
	dfs(P,0);int Q=0;
	for(int i=1;i<=n;++i)Q=dep[i]>dep[Q]?i:Q;
	for(int i=Q;i;i=fa[i])mrk[i]=++len;
	for(int i=1;i<=n;++i)
		if(!mrk[i]){
			if(!mrk[fa[i]])return puts("-1"),0;
			++t[mrk[fa[i]]];
		}
	for(int i=1;i<=len;++i)
		if(t[i]!=t[len-i+1]){
			if(t[i]>t[len-i+1])
				reverse(t+1,t+len+1);
			break;
		}
	for(int i=1,j=0;i<=len;++i){
		for(int k=j+2;k<=j+t[i]+1;++k)printf("%d ",k);
		printf("%d ",j+1);j+=t[i]+1;
	}
	puts("");return 0;
}
