#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define REP(i,n) for(int i=(0);i<(n);i++)
#define fi first
#define se second
#define pb push_back
#define mp make_pair
using namespace std;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef long long ll;
typedef double ld;
 
template<class T> inline void read(T &x){
	int f=0;x=0;char ch=getchar();
	for(;!isdigit(ch);ch=getchar())f|=(ch=='-');
	for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';
	if(f)x=-x;
}

const int N=100005;
int fa[N],d[N],a[N],cur[N],c[N],tmp[N],ans[N];
int num,pos,cnt,s,t,n,m,u,v;
vi e[N];

void dfs(int u){
	for(auto v:e[u])
		if(v!=fa[u]){
			d[v]=d[u]+1;
			fa[v]=u,dfs(v);
		}
}

int cmp(int *a,int *b){
	rep(i,1,cnt)
		if(a[i]!=b[i])return a[i]<b[i];
	return 0;
}

int main(){
	read(n);
	rep(i,1,n-1){
		read(u),read(v);
		e[u].pb(v),e[v].pb(u);
	}
	fa[1]=0,d[1]=1,dfs(1);
	s=max_element(d+1,d+n+1)-d;
	fa[s]=0,d[s]=1,dfs(s);
	t=max_element(d+1,d+n+1)-d;
	a[++cnt]=t;
	while(a[cnt]!=s){
		int t=fa[a[cnt]];
		a[++cnt]=t;
	}
	rep(i,1,cnt)cur[a[i]]=i;
	rep(i,1,n)
		if(!cur[i]){
			if(a[cur[fa[i]]]!=fa[i])
				return puts("-1"),0;
			cur[i]=cur[fa[i]];
			++c[cur[i]];
		}
	memcpy(tmp,c,sizeof c);
	reverse(tmp+1,tmp+cnt+1);
	if(cmp(tmp,c))memcpy(c,tmp,sizeof tmp);
	rep(i,1,cnt){
		int tmp=++num;
		rep(j,1,c[i])
			ans[pos+j]=++num;
		pos+=c[i]+1;
		ans[pos]=tmp;
	}
	rep(i,1,n)printf("%d ",ans[i]);
	return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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>
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define REP(i,n) for(int i=(0);i<(n);i++)
#define fi first
#define se second
#define pb push_back
#define mp make_pair
using namespace std;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef long long ll;
typedef double ld;
 
template<class T> inline void read(T &x){
	int f=0;x=0;char ch=getchar();
	for(;!isdigit(ch);ch=getchar())f|=(ch=='-');
	for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';
	if(f)x=-x;
}

const int N=100005;
int fa[N],d[N],a[N],cur[N],c[N],tmp[N],ans[N];
int num,pos,cnt,s,t,n,m,u,v;
vi e[N];

void dfs(int u){
	for(auto v:e[u])
		if(v!=fa[u]){
			d[v]=d[u]+1;
			fa[v]=u,dfs(v);
		}
}

int cmp(int *a,int *b){
	rep(i,1,cnt)
		if(a[i]!=b[i])return a[i]<b[i];
	return 0;
}

int main(){
	read(n);
	rep(i,1,n-1){
		read(u),read(v);
		e[u].pb(v),e[v].pb(u);
	}
	fa[1]=0,d[1]=1,dfs(1);
	s=max_element(d+1,d+n+1)-d;
	fa[s]=0,d[s]=1,dfs(s);
	t=max_element(d+1,d+n+1)-d;
	a[++cnt]=t;
	while(a[cnt]!=s){
		int t=fa[a[cnt]];
		a[++cnt]=t;
	}
	rep(i,1,cnt)cur[a[i]]=i;
	rep(i,1,n)
		if(!cur[i]){
			if(a[cur[fa[i]]]!=fa[i])
				return puts("-1"),0;
			cur[i]=cur[fa[i]];
			++c[cur[i]];
		}
	memcpy(tmp,c,sizeof c);
	reverse(tmp+1,tmp+cnt+1);
	if(cmp(tmp,c))memcpy(c,tmp,sizeof tmp);
	rep(i,1,cnt){
		int tmp=++num;
		rep(j,1,c[i])
			ans[pos+j]=++num;
		pos+=c[i]+1;
		ans[pos]=tmp;
	}
	rep(i,1,n)printf("%d ",ans[i]);
	return 0;
}
```