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 sz(v) ((int)(v).size())
using namespace std;
using pi = pair<int, int>;
using lint = long long;
const int MAXN = 100005;

int n;
vector<int> gph[MAXN];
int ret[MAXN], par[MAXN], chk[MAXN];

pi dfs(int x, int p){
	pi ret(0, x);
	for(auto &i : gph[x]){
		if(i == p) continue;
		par[i] = x;
		auto k = dfs(i, x);
		k.first++;
		ret = max(ret, k);
	}
	return ret;
}

int main(){
	scanf("%d",&n);
	for(int i=1; i<n; i++){
		int s, e; scanf("%d %d",&s,&e);
		gph[s].push_back(e);
		gph[e].push_back(s);
	}
	int st = dfs(1, -1).second;
	int ed = dfs(st, -1).second;
	for(int j=ed; ; j=par[j]){
		chk[j] = j;
		if(j == st) break;
	}
	for(int i=1; i<=n; i++){
		if(!chk[i]){
			bool ok = 0;
			for(auto &j : gph[i]){
				if(chk[j]) ok = true;
			}
			if(!ok){
				puts("-1");
				return 0;
			}
		}
	}
	vector<int> v;
	for(int j=ed; ; j=par[j]){
		int cnt = 1;
		for(auto &k : gph[j]){
			if(!chk[k]) cnt++;
		}
		v.push_back(cnt);
		if(j == st) break;
	}
	v = min(v, vector<int>(v.rbegin(), v.rend()));
	int sum = 0;
	int p = 0;
	for(auto &i : v){
		for(int j=sum+2; j<=sum+i; j++){
			ret[++p] = j;
		}
		ret[++p] = sum + 1;
		sum += i;
	}
	for(int j=1; j<=n; j++) printf("%d ", ret[j]);
}
