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;

int read(){
	int a = 0; char c = getchar();
	while(!isdigit(c)) c = getchar();
	while(isdigit(c)){a = a * 10 + c - 48; c = getchar();}
	return a;
}

const int _ = 1e5 + 7;
struct Edge{int end , upEd;}Ed[_ << 1];
int head[_] , cntEd , N , mx , id;

void addEd(int a , int b){Ed[++cntEd] = (Edge){b , head[a]}; head[a] = cntEd;}

int pre[_]; bool vis[_];
void dfs(int x , int p , int l){
	if(l > mx){mx = l; id = x;}
	pre[x] = p;
	for(int i = head[x] ; i ; i = Ed[i].upEd)
		if(Ed[i].end != p) dfs(Ed[i].end , x , l + 1);
}

int main(){
	N = read();
	for(int i = 1 ; i < N ; ++i){int x = read() , y = read(); addEd(x , y); addEd(y , x);}
	dfs(1 , 0 , 0); int p = id; mx = 0; dfs(p , 0 , 0); int q = id;
	vis[id] = 1; while(id) vis[id = pre[id]] = 1;
	for(int i = 1 ; i <= N ; ++i) if(!vis[i] && !vis[pre[i]]){puts("-1"); return 0;}
	vector < int > node; deque < int > ans1 , ans2; int cnt = 0;
	for(int t = q ; t ; t = pre[t]) node.push_back(t);
	
	for(auto q : node){
		int now = ++cnt , sz = 0;
		for(int i = head[q] ; i ; i = Ed[i].upEd) sz += !vis[Ed[i].end];
		for(int i = now + 1 ; i <= now + sz ; ++i) ans1.push_back(i);
		ans1.push_back(now); cnt += sz;
	}
	
	reverse(node.begin() , node.end()); cnt = 0;
	for(auto q : node){
		int now = ++cnt , sz = 0;
		for(int i = head[q] ; i ; i = Ed[i].upEd) sz += !vis[Ed[i].end];
		for(int i = now + 1 ; i <= now + sz ; ++i) ans2.push_back(i);
		ans2.push_back(now); cnt += sz;
	}

	for(int i = 0 ; i < ans1.size() ; ++i)
		if(ans1[i] < ans2[i]) break;
		else if(ans1[i] > ans2[i]){swap(ans1 , ans2); break;}
	for(auto t : ans1) cout << t << ' ';
	
	return 0;
}