//Love and Freedom.
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#define ll long long
#define inf 20021225
#define N 100010 
using namespace std;
int read()
{
	int s=0,t=1; char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')	t=-1; ch=getchar();}
	while(ch>='0' && ch<='9')	s=s*10+ch-'0',ch=getchar();
	return s*t;
}
struct edge{int to,lt;}e[N<<1]; int in[N],cnt;
void add(int x,int y)
{
	e[++cnt].to=y; e[cnt].lt=in[x]; in[x]=cnt;
	e[++cnt].to=x; e[cnt].lt=in[y]; in[y]=cnt;
}
int f[N],d[N],sz[N],qwq[N],n,p,q,len,gg,nxt[N];
void dfs(int x,int fr)
{
	d[x]=x; f[x]=1;
	for(int i=in[x];i;i=e[i].lt)
	{
		int y=e[i].to; if(y==fr)	continue; dfs(y,x);
		if(f[y]+f[x]>len) len=f[y]+f[x],p=d[x],q=d[y];
		if(f[y]+1>f[x]) f[x]=f[y]+1,d[x]=d[y];
	}
}
void dfs2(int x,int fr)
{
	sz[x]=qwq[x]=1; if(gg)	return; int dd=0;
	for(int i=in[x];i;i=e[i].lt)
	{
		int y=e[i].to; if(y==fr)	continue;
		dfs2(y,x); if(sz[y]>1)	dd++,nxt[x]=y; else qwq[x]++;
		sz[x]+=sz[y]; 
	}
	if(dd>1)	gg=1;
}
int a[N],b[N],g[N];
int main()
{
	n=read(); for(int i=1;i<n;i++)	add(read(),read());
	dfs(1,0); dfs2(p,0); int tot=0; if(gg)	return puts("-1"),0;
	for(int i=p;i;i=nxt[i])	g[++tot]=qwq[i];
	g[++tot]=1; int ss=0; g[tot-1]--;
	for(int i=1;i<=tot;i++)
	{
		++ss; a[ss+g[i]-1]=ss; 
		for(int x=2;x<=g[i];x++)	++ss,a[ss-1]=ss;
	}
	reverse(g+1,g+tot+1); ss=0;
	for(int i=1;i<=tot;i++)
	{
		++ss; b[ss+g[i]-1]=ss;
		for(int x=2;x<=g[i];x++)	++ss,b[ss-1]=ss;
	}
	int owo=1;
	for(int i=1;i<=n;i++)	if(a[i]!=b[i]){owo=a[i]<b[i]; break;}
	if(owo)	for(int i=1;i<=n;i++)	printf("%d ",a[i]);
	else	for(int i=1;i<=n;i++)	printf("%d ",b[i]);
	return 0;
}
### 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
//Love and Freedom.
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#define ll long long
#define inf 20021225
#define N 100010 
using namespace std;
int read()
{
	int s=0,t=1; char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')	t=-1; ch=getchar();}
	while(ch>='0' && ch<='9')	s=s*10+ch-'0',ch=getchar();
	return s*t;
}
struct edge{int to,lt;}e[N<<1]; int in[N],cnt;
void add(int x,int y)
{
	e[++cnt].to=y; e[cnt].lt=in[x]; in[x]=cnt;
	e[++cnt].to=x; e[cnt].lt=in[y]; in[y]=cnt;
}
int f[N],d[N],sz[N],qwq[N],n,p,q,len,gg,nxt[N];
void dfs(int x,int fr)
{
	d[x]=x; f[x]=1;
	for(int i=in[x];i;i=e[i].lt)
	{
		int y=e[i].to; if(y==fr)	continue; dfs(y,x);
		if(f[y]+f[x]>len) len=f[y]+f[x],p=d[x],q=d[y];
		if(f[y]+1>f[x]) f[x]=f[y]+1,d[x]=d[y];
	}
}
void dfs2(int x,int fr)
{
	sz[x]=qwq[x]=1; if(gg)	return; int dd=0;
	for(int i=in[x];i;i=e[i].lt)
	{
		int y=e[i].to; if(y==fr)	continue;
		dfs2(y,x); if(sz[y]>1)	dd++,nxt[x]=y; else qwq[x]++;
		sz[x]+=sz[y]; 
	}
	if(dd>1)	gg=1;
}
int a[N],b[N],g[N];
int main()
{
	n=read(); for(int i=1;i<n;i++)	add(read(),read());
	dfs(1,0); dfs2(p,0); int tot=0; if(gg)	return puts("-1"),0;
	for(int i=p;i;i=nxt[i])	g[++tot]=qwq[i];
	g[++tot]=1; int ss=0; g[tot-1]--;
	for(int i=1;i<=tot;i++)
	{
		++ss; a[ss+g[i]-1]=ss; 
		for(int x=2;x<=g[i];x++)	++ss,a[ss-1]=ss;
	}
	reverse(g+1,g+tot+1); ss=0;
	for(int i=1;i<=tot;i++)
	{
		++ss; b[ss+g[i]-1]=ss;
		for(int x=2;x<=g[i];x++)	++ss,b[ss-1]=ss;
	}
	int owo=1;
	for(int i=1;i<=n;i++)	if(a[i]!=b[i]){owo=a[i]<b[i]; break;}
	if(owo)	for(int i=1;i<=n;i++)	printf("%d ",a[i]);
	else	for(int i=1;i<=n;i++)	printf("%d ",b[i]);
	return 0;
}
```