#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<ll,ll> pi;
typedef vector <ll> vi;
typedef vector <pi> vpi;
#define f first
#define s second
#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)
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define lbd(x, y) lower_bound(all(x), y)
#define ubd(x, y) upper_bound(all(x), y)
#define aFOR(i,x) for (auto i: x)
#define mem(x,i) memset(x,i,sizeof x)
#define fast ios_base::sync_with_stdio(false),cin.tie(0)
#define maxn 100001

int N,st,e;
vi adj[maxn];
int par[maxn];
int mxdist = 0,mxnode;

void dfs(int x,int p,int d){
	par[x] = p;
	if (d > mxdist){
		mxnode = x;
		mxdist = d;
	}
	
	aFOR(i,adj[x]){
		if (i != p){
			dfs(i,x,d + 1);
		}
	}
}
bool inbody[maxn];

int main(){
	fast;
	
	cin>>N;
	
	FOR(i,1,N-1){
		int a,b;
		cin>>a>>b;
		adj[a].pb(b);
		adj[b].pb(a);
	}
	
	dfs(1,-1,0);
	mxdist = 0;
	
	st = mxnode;
	dfs(mxnode,-1,0);
	e = mxnode;
	
	vi body;
	
	int a = e;
	
	while (a != st){
		inbody[a] = 1;
		body.pb(a);
		a = par[a];
	}
	inbody[st] = 1;
	body.pb(st);
	
	reverse(all(body));
	
	FOR(i,1,N){
		if (!inbody[i]){
			if (adj[i].size() != 1){
				cout<<-1;
				return 0;
			}
		}
	}
	vi res,ans;
	
	int count = 1;
	
	aFOR(i,body){
		int num = 0;
		aFOR(j,adj[i]) if (!inbody[j]){
			num++;
		}
		res.pb(count + num);
		aFOR(j,adj[i]) if (!inbody[j]){
			res.pb(count);
			count++;
		}	
		count++;

	}
	vi res2;
	
	FOR(i,1,N) res2.pb(0);
	FOR(i,0,N-1){
		res2[res[i] - 1] = i + 1;
	}
	
	ans = res2;

	reverse(all(body));
	res.clear();
	
	count = 1;
	
	aFOR(i,body){
		int num = 0;
		aFOR(j,adj[i]) if (!inbody[j]){
			num++;
		}
		res.pb(count + num);
		aFOR(j,adj[i]) if (!inbody[j]){
			res.pb(count);
			count++;
		}	
		count++;

	}
	
	res2.clear();
	
	FOR(i,1,N) res2.pb(0);
	FOR(i,0,N-1){
		res2[res[i] - 1] = i + 1;
	}
	
	ans = min(ans,res2);
	aFOR(i,ans){
		cout<<i<<' ';
	}
}


### Prompt

Your task is to create a cpp solution to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef pair<ll,ll> pi;
typedef vector <ll> vi;
typedef vector <pi> vpi;
#define f first
#define s second
#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)
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define lbd(x, y) lower_bound(all(x), y)
#define ubd(x, y) upper_bound(all(x), y)
#define aFOR(i,x) for (auto i: x)
#define mem(x,i) memset(x,i,sizeof x)
#define fast ios_base::sync_with_stdio(false),cin.tie(0)
#define maxn 100001

int N,st,e;
vi adj[maxn];
int par[maxn];
int mxdist = 0,mxnode;

void dfs(int x,int p,int d){
	par[x] = p;
	if (d > mxdist){
		mxnode = x;
		mxdist = d;
	}
	
	aFOR(i,adj[x]){
		if (i != p){
			dfs(i,x,d + 1);
		}
	}
}
bool inbody[maxn];

int main(){
	fast;
	
	cin>>N;
	
	FOR(i,1,N-1){
		int a,b;
		cin>>a>>b;
		adj[a].pb(b);
		adj[b].pb(a);
	}
	
	dfs(1,-1,0);
	mxdist = 0;
	
	st = mxnode;
	dfs(mxnode,-1,0);
	e = mxnode;
	
	vi body;
	
	int a = e;
	
	while (a != st){
		inbody[a] = 1;
		body.pb(a);
		a = par[a];
	}
	inbody[st] = 1;
	body.pb(st);
	
	reverse(all(body));
	
	FOR(i,1,N){
		if (!inbody[i]){
			if (adj[i].size() != 1){
				cout<<-1;
				return 0;
			}
		}
	}
	vi res,ans;
	
	int count = 1;
	
	aFOR(i,body){
		int num = 0;
		aFOR(j,adj[i]) if (!inbody[j]){
			num++;
		}
		res.pb(count + num);
		aFOR(j,adj[i]) if (!inbody[j]){
			res.pb(count);
			count++;
		}	
		count++;

	}
	vi res2;
	
	FOR(i,1,N) res2.pb(0);
	FOR(i,0,N-1){
		res2[res[i] - 1] = i + 1;
	}
	
	ans = res2;

	reverse(all(body));
	res.clear();
	
	count = 1;
	
	aFOR(i,body){
		int num = 0;
		aFOR(j,adj[i]) if (!inbody[j]){
			num++;
		}
		res.pb(count + num);
		aFOR(j,adj[i]) if (!inbody[j]){
			res.pb(count);
			count++;
		}	
		count++;

	}
	
	res2.clear();
	
	FOR(i,1,N) res2.pb(0);
	FOR(i,0,N-1){
		res2[res[i] - 1] = i + 1;
	}
	
	ans = min(ans,res2);
	aFOR(i,ans){
		cout<<i<<' ';
	}
}


```