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;

#define FAST ios_base::sync_with_stdio(false); cin.tie(0);
#define pb push_back
#define eb emplace_back
#define ins insert
#define f first
#define s second
#define cbr cerr << "hi\n"
#define mmst(x, v) memset((x), v, sizeof ((x)))
#define siz(x) ll(x.size())
#define all(x) (x).begin(), (x).end()
#define lbd(x,y) (lower_bound(all(x),y)-x.begin())
#define ubd(x,y) (upper_bound(all(x),y)-x.begin())
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());    //can be used by calling rng() or shuffle(A, A+n, rng)
inline long long rand(long long x, long long y) { return (rng() % (y+1-x)) + x; } //inclusivesss
string inline to_string(char c) {string s(1,c);return s;} template<typename T> inline T gcd(T a,T b){ return a==0?llabs(b):gcd(b%a,a); }

typedef long long ll; 
typedef long double ld;
#define FOR(i,s,e) for(ll i=s;i<=ll(e);++i)
#define DEC(i,s,e) for(ll i=s;i>=ll(e);--i)
typedef pair<ll,ll>pi; typedef pair<ll,pi>spi; typedef pair<pi,pi>dpi;

#define LLINF ((long long)1e18)
#define INF int(1e9+1e6)
#define MAXN (100006)
ll n,dist[MAXN],p[MAXN];
vector<int>v[MAXN];
bitset<MAXN> in;
int main(){
	FAST
	cin>>n;
	FOR(i,0,n-2){
		ll a,b;cin>>a>>b;
		v[a].eb(b), v[b].eb(a);
	}
	function<void(ll,ll)>dfs=[&](ll x,ll par){
		p[x]=par;
		for(auto i:v[x]) if(i^par) {
			dist[i]=dist[x]+1, dfs(i,x);
		}
	};
	dfs(1,1);
	ll mx = 1;
	FOR(i,1,n) if(dist[i]>dist[mx]) mx=i;
	dist[mx]=0, dfs(mx,mx);
	ll mx2=mx;
	FOR(i,1,n) if(dist[i]>dist[mx2]) mx2=i;
	vector<ll> path;
	for(ll i=mx2; ; i=p[i]){
		in[i]=1;
		path.eb(siz(v[i])-1-ll(i!=mx2&&i!=mx));
		if(i==mx) break;
	}
	FOR(i,1,n) if(!in[i] && siz(v[i]) > 1) { cout<<"-1\n"; return 0; }
	bool rev = 0;
	for(ll co=0, co1=siz(path)-1; co < co1; ++ co, -- co1){
		if(path[co] > path[co1]) { rev=1; break; }
		else if(path[co] < path[co1]) break;
	}
	if(rev) reverse(all(path));
	ll co=1;
	for(auto i:path){
		ll myval=co++;
		while(i--) cout<<co++<<' ';
		cout<<myval<<' ';
	}cout<<'\n';
}
