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>
#define pb emplace_back
#define N 100010
using namespace std;
int d[N],n,x,y,a[N],na;
vector<int> e[N];
void jb(int x,int y){e[x].pb(y);d[x]++;}
void get(int x,int fa){
	a[++na]=e[x].size();
	for (auto y:e[x])
		if (y!=fa&&e[y].size()>1)get(y,x);
}
bool ch(){
	for (int i=0;i<=na;i++){if (a[na-i]>a[i]) return 0;if (a[i]>a[na-i])return 1;}
	return 0;
}
int main(){
	cin>>n;
	for (int i=1;i<n;i++){cin>>x>>y;jb(x,y);jb(y,x);}
	for (int i=1;i<=n;i++)if (e[i].size()==1)d[e[i][0]]--;
	for (int i=1;i<=n;i++)if (d[i]>2){puts("-1");return 0;}
	if (n==2){puts("1 2");return 0;}
	for (int i=1;i<=n;i++)if (e[i].size()>1&&d[i]<2){get(i,0);break;}
	for (int i=1;i<=na;i++)a[i]-=2;
	na++;
	if (ch())reverse(a,a+na+1);
	for (int i=0,su=0;i<=na;i++){
		for (int j=su+1;j<=su+a[i];j++)cout<<j+1<<' ';
		su+=a[i]+1;cout<<su-a[i]<<' ';
	}
	return 0;
}