#include <cstdio>
#include <algorithm>

using namespace std;

int edge[300000],nxt[300000],first[300000];
int c1[300000],c2[300000],d1[300000],d2[300000],e[300000];
int i,j,k,n,p,t,t1,t2,x,y,sum_edge;

inline void addedge(int x,int y)
{
	sum_edge++,edge[sum_edge]=y,nxt[sum_edge]=first[x],first[x]=sum_edge; 
	return;
}

int main()
{
	scanf("%d",&n);
	if (n==2)
		return printf("1 2"),0;
	if (n==3)
		return printf("1 2 3"),0;
	for (i=1;i<n;i++)
	{
		scanf("%d%d",&x,&y);
		addedge(x,y),addedge(y,x);
	}
	for (i=1;i<=n;i++)
		for (j=first[i];j!=0;j=nxt[j])
			if (nxt[first[edge[j]]])
				if (! d1[i])
					d1[i]=edge[j];
				else
					if (! d2[i])
						d2[i]=edge[j];
					else
						return puts("-1"),0;
			else
				e[i]++;
	for (i=1;i<=n;i++)
		if ((d1[i]) && (d1[d1[i]]!=i) && (d2[d1[i]]!=i))
			if (! d1[d1[i]])
			{
				d1[d1[i]]=i;
				e[d1[i]]--;
				if (! t1)
					t1=i;
				else
					t2=i;
			}
			else
				if (! d2[d1[i]])
				{
					d2[d1[i]]=i;
					e[d1[i]]--;
					if (! t1)
						t1=i;
					else
						t2=i;
				}
	for (k=0,p=0,i=t1;i!=t2;)
	{
		k++,c1[k]=e[i]+1;
		if (d1[i]==p)
			p=i,i=d2[i];
		else
			p=i,i=d1[i];
	}
	k++,c1[k]=e[i]+1;
	swap(t1,t2);
	for (k=0,p=0,i=t1;i!=t2;)
	{
		k++,c2[k]=e[i]+1;
		if (d1[i]==p)
			p=i,i=d2[i];
		else
			p=i,i=d1[i];
	}
	k++,c2[k]=e[i]+1;
	t1=1,t2=0;
	for (i=1;i<=k;i++)
	{
		if (c1[i]<c2[i])
		{
			t1=1,t2=0;
			break;
		}
		if (c2[i]<c1[i])
		{
			t1=0,t2=1;
			break;
		}
	}
	t=0;
	for (i=1;i<=k;i++)
		if (t1)
		{
			for (j=2;j<=c1[i];j++)
				printf("%d ",t+j);
			printf("%d ",t+1);
			t=t+c1[i];
		}
		else
		{
			for (j=2;j<=c2[i];j++)
				printf("%d ",t+j);
			printf("%d ",t+1);
			t=t+c2[i];
		}
	return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 <cstdio>
#include <algorithm>

using namespace std;

int edge[300000],nxt[300000],first[300000];
int c1[300000],c2[300000],d1[300000],d2[300000],e[300000];
int i,j,k,n,p,t,t1,t2,x,y,sum_edge;

inline void addedge(int x,int y)
{
	sum_edge++,edge[sum_edge]=y,nxt[sum_edge]=first[x],first[x]=sum_edge; 
	return;
}

int main()
{
	scanf("%d",&n);
	if (n==2)
		return printf("1 2"),0;
	if (n==3)
		return printf("1 2 3"),0;
	for (i=1;i<n;i++)
	{
		scanf("%d%d",&x,&y);
		addedge(x,y),addedge(y,x);
	}
	for (i=1;i<=n;i++)
		for (j=first[i];j!=0;j=nxt[j])
			if (nxt[first[edge[j]]])
				if (! d1[i])
					d1[i]=edge[j];
				else
					if (! d2[i])
						d2[i]=edge[j];
					else
						return puts("-1"),0;
			else
				e[i]++;
	for (i=1;i<=n;i++)
		if ((d1[i]) && (d1[d1[i]]!=i) && (d2[d1[i]]!=i))
			if (! d1[d1[i]])
			{
				d1[d1[i]]=i;
				e[d1[i]]--;
				if (! t1)
					t1=i;
				else
					t2=i;
			}
			else
				if (! d2[d1[i]])
				{
					d2[d1[i]]=i;
					e[d1[i]]--;
					if (! t1)
						t1=i;
					else
						t2=i;
				}
	for (k=0,p=0,i=t1;i!=t2;)
	{
		k++,c1[k]=e[i]+1;
		if (d1[i]==p)
			p=i,i=d2[i];
		else
			p=i,i=d1[i];
	}
	k++,c1[k]=e[i]+1;
	swap(t1,t2);
	for (k=0,p=0,i=t1;i!=t2;)
	{
		k++,c2[k]=e[i]+1;
		if (d1[i]==p)
			p=i,i=d2[i];
		else
			p=i,i=d1[i];
	}
	k++,c2[k]=e[i]+1;
	t1=1,t2=0;
	for (i=1;i<=k;i++)
	{
		if (c1[i]<c2[i])
		{
			t1=1,t2=0;
			break;
		}
		if (c2[i]<c1[i])
		{
			t1=0,t2=1;
			break;
		}
	}
	t=0;
	for (i=1;i<=k;i++)
		if (t1)
		{
			for (j=2;j<=c1[i];j++)
				printf("%d ",t+j);
			printf("%d ",t+1);
			t=t+c1[i];
		}
		else
		{
			for (j=2;j<=c2[i];j++)
				printf("%d ",t+j);
			printf("%d ",t+1);
			t=t+c2[i];
		}
	return 0;
}

```