#include<bits/stdc++.h>
#define ll long long
#define ull unsigned ll
#define uint unsigned
#define pii pair<int,int>
#define pll pair<ll,ll>
#define PB push_back
#define fi first
#define se second
#define For(i,j,k) for (int i=(int)(j);i<=(int)(k);i++)
#define Rep(i,j,k) for (int i=(int)(j);i>=(int)(k);i--)
#define CLR(a,v) memset(a,v,sizeof(a));
#define CPY(a,b) memcpy(a,b,sizeof(a));
using namespace std;
const int N=100005;
struct edge{
	int to,next;
}e[N*2];
int head[N],tot,n;
int deg[N],q[N],DS[N],DT[N];
int vis[N],ans1[N],ans2[N];
void add(int x,int y){
	e[++tot]=(edge){y,head[x]};
	head[x]=tot;
}
int bfs(int S,int *dis){
	For(i,1,n) dis[i]=-1;
	int h=0,t=1;
	q[1]=S; dis[S]=0;
	while (h!=t){
		int x=q[++h];
		for (int i=head[x];i;i=e[i].next)
			if (dis[e[i].to]==-1){
				dis[e[i].to]=dis[x]+1;
				q[++t]=e[i].to;
			}
	}
	return q[t];
}
void work(int *q,int l,int *ans){
	*ans=0;
	int S=0;
	For(i,0,l){
		int v=deg[q[i]];
		v-=(i!=0)+(i!=l);
		For(j,1,v) ans[++*ans]=S+j+1;
		ans[++*ans]=S+1;
		S+=v+1;
	}
}
int main(){
	scanf("%d",&n);
	For(i,1,n-1){
		int x,y;
		scanf("%d%d",&x,&y);
		add(x,y); add(y,x);
		++deg[x]; ++deg[y];
	}
	int S=bfs(1,DS);
	int T=bfs(S,DS);
	bfs(T,DT);
	For(i,1,n)
		if (DS[i]+DT[i]==DS[T]){
			q[DS[i]]=i;
			for (int j=head[i];j;j=e[j].next)
				vis[e[j].to]=1;
		}
	For(i,1,n)
		if (!vis[i])
			return puts("-1"),0;
	work(q,DS[T],ans1);
	reverse(q,q+DS[T]+1);
	work(q,DS[T],ans2);
	int fl=0;
	For(i,1,n){
		if (!fl&&ans1[i]!=ans2[i])
			fl=(ans1[i]<ans2[i]?1:-1);
		printf("%d ",fl==1?ans1[i]:ans2[i]);
	}
}
### Prompt

Generate 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>
#define ll long long
#define ull unsigned ll
#define uint unsigned
#define pii pair<int,int>
#define pll pair<ll,ll>
#define PB push_back
#define fi first
#define se second
#define For(i,j,k) for (int i=(int)(j);i<=(int)(k);i++)
#define Rep(i,j,k) for (int i=(int)(j);i>=(int)(k);i--)
#define CLR(a,v) memset(a,v,sizeof(a));
#define CPY(a,b) memcpy(a,b,sizeof(a));
using namespace std;
const int N=100005;
struct edge{
	int to,next;
}e[N*2];
int head[N],tot,n;
int deg[N],q[N],DS[N],DT[N];
int vis[N],ans1[N],ans2[N];
void add(int x,int y){
	e[++tot]=(edge){y,head[x]};
	head[x]=tot;
}
int bfs(int S,int *dis){
	For(i,1,n) dis[i]=-1;
	int h=0,t=1;
	q[1]=S; dis[S]=0;
	while (h!=t){
		int x=q[++h];
		for (int i=head[x];i;i=e[i].next)
			if (dis[e[i].to]==-1){
				dis[e[i].to]=dis[x]+1;
				q[++t]=e[i].to;
			}
	}
	return q[t];
}
void work(int *q,int l,int *ans){
	*ans=0;
	int S=0;
	For(i,0,l){
		int v=deg[q[i]];
		v-=(i!=0)+(i!=l);
		For(j,1,v) ans[++*ans]=S+j+1;
		ans[++*ans]=S+1;
		S+=v+1;
	}
}
int main(){
	scanf("%d",&n);
	For(i,1,n-1){
		int x,y;
		scanf("%d%d",&x,&y);
		add(x,y); add(y,x);
		++deg[x]; ++deg[y];
	}
	int S=bfs(1,DS);
	int T=bfs(S,DS);
	bfs(T,DT);
	For(i,1,n)
		if (DS[i]+DT[i]==DS[T]){
			q[DS[i]]=i;
			for (int j=head[i];j;j=e[j].next)
				vis[e[j].to]=1;
		}
	For(i,1,n)
		if (!vis[i])
			return puts("-1"),0;
	work(q,DS[T],ans1);
	reverse(q,q+DS[T]+1);
	work(q,DS[T],ans2);
	int fl=0;
	For(i,1,n){
		if (!fl&&ans1[i]!=ans2[i])
			fl=(ans1[i]<ans2[i]?1:-1);
		printf("%d ",fl==1?ans1[i]:ans2[i]);
	}
}
```