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 maxn = 123456;
vector<int> e[maxn];
int dis[maxn], par[maxn], good[maxn];
void dfs(int u, int v){
	par[u] = v;
	for(int nxt : e[u]){
		if(nxt == v) continue;
		dis[nxt] = dis[u] + 1;
		dfs(nxt, u);
	}
}
int main(){
	int n; cin >> n;
	for(int i = 0; i < n - 1; i++){
		int u, v; cin >> u >> v;
		e[u].push_back(v);
		e[v].push_back(u);
	}
	auto bad = [](){
		cout << -1 << endl;
		exit(0);
	};
	dis[1] = 0; dfs(1, 0);
	int mx = max_element(dis + 1, dis + 1 + n) - dis;
	dis[mx] = 0; dfs(mx, 0);
	int mn = max_element(dis + 1, dis + 1 + n) - dis;
	{
		int tail = mn;
		while(tail){
			good[tail] = 1;
			tail = par[tail];
		}
	}
	for(int i = 1; i <= n; i++){
		int sum = good[i];
		for(int v : e[i]) sum += good[v];
		if(sum == 0) bad();
	}
	auto get = [&](int head){
		vector<int> path;
		int id = 1, lhead = -1;
		while(head != -1){
			int nxt = -1, sum = 0;
			for(int v : e[head]){
				if(v == lhead) continue;
				if(good[v]) nxt = v;
				else {
					sum++;
				}
			}
			int me = id++;
			for(int j = 0; j < sum; j++) path.push_back(id++);
			path.push_back(me);
			lhead = head;
			head = nxt;
		}
		return path;
	};
	vector<int> res = min(get(mn), get(mx));
	for(int i = 0; i < n; i++) cout << res[i] << " "; cout << endl;
	return 0;
}
