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;
const int N=1e5+5;
int n,x,y,i,j,dd[N],p[N];
vector<int>e[N],p0,p1;
int nxt(int x,int fa){for(int y:e[x])if(e[y].size()>1 && y!=fa)return y;return 0;}
int main(){
	scanf("%d",&n);for(i=1;i<n;++i)scanf("%d%d",&x,&y),e[x].push_back(y),e[y].push_back(x);
	if(n==2){puts("1 2");return 0;}
	for(i=1;i<=n;++i)for(int x:e[i])dd[i]+=e[x].size()>1;
	for(i=1;i<=n;++i)if(dd[i]>2){puts("-1");return 0;}
	for(i=1;i<=n;++i)if(dd[i]<2 && e[i].size()>1){
		auto&v=p0.empty()?p0:p1;int lst=0;
		for(x=i;x;v.push_back(x),y=x,x=nxt(x,lst),lst=y);
		for(int&x:v)x=e[x].size()-2;
	}
	if(!p1.empty() && p0>p1)p0=p1;
	p[1]=1;p[n]=n;x=y=1;
	for(i=0;i<p0.size();++i){
		p[x+p0[i]+1]=++y;
		for(j=0;j<p0[i];++j)p[x+j+1]=++y;x+=p0[i]+1;
	}
	for(i=1;i<=n;++i)printf("%d%c",p[i],i==n?'\n':' ');
	return 0;
}