#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<queue>
#include<bitset>
#include<complex>
using namespace std;
template<typename __T>
inline void read(__T &x)
{
    x=0;
    int f=1;char c=getchar();
    while(!isdigit(c)){if(c=='-')	f=-1;c=getchar();}
    while(isdigit(c))	{x=x*10+c-'0';c=getchar();}
    x*=f;
}
int box[100005],las[200005],edv[200005],cnt=0;
void adde(int u,int v)
{
	las[++cnt]=box[u];
	box[u]=cnt;
	edv[cnt]=v;
}
int dep[100005],dfsin[100005],dfn=0;
int tot=0;
int pre[100005];
void dfs(int now,int fa)
{
	pre[now]=fa;
	dfsin[now]=++dfn;
	for(int i=box[now];i;i=las[i])
		if(edv[i]!=fa)
		{
			dep[edv[i]]=dep[now]+1;
			dfs(edv[i],now);
		}
}
int S,T;
int n;
int ans[100005];
int ans2[100005];
int deg[100005];
bool fg=0;
bool onl[100005];
void solve(int now,int fa)
{
	dfsin[now]=++dfn;
	if(!onl[now])
	{
		if(deg[now]>1)	fg=1;
		return;
	}
	if(now==T)
	{
		ans[tot++]=dfsin[now];
		return;
	}
	int s=0;
	for(int i=box[now];i;i=las[i])
		if(edv[i]!=fa)
		{
			int v=edv[i];
			if(!onl[v])
			{
				solve(v,now);
				ans[tot++]=dfsin[v];
			}
			else
				s=v;
		}
	ans[tot++]=dfsin[now];
	solve(s,now);
}
int main()
{
	read(n);
	int a,b;
	for(int i=1;i<n;i++)
	{
		read(a);
		read(b);
		deg[a]++;
		deg[b]++;
		adde(a,b);
		adde(b,a);
	}
	dfs(1,0);
	for(int i=1;i<=n;i++)
		if(dep[i]>dep[S])	S=i;
	dfn=0;
	dfs(S,0);
	for(int i=1;i<=n;i++)
		if(dep[i]>dep[T])	T=i;
	for(int i=T;i;i=pre[i])	onl[i]=1;
	dfn=tot=0;
	solve(S,0);
	if(fg)
	{
		printf("-1\n");
		return 0;
	}
	for(int i=0;i<n;i++)
		ans2[i]=ans[i];
	swap(S,T);
	tot=dfn=0;
	solve(S,0);
	for(int i=0;i<n;i++)
	{
		if(ans[i]<ans2[i])
		{
			for(int i=0;i<n;i++)
				printf("%d ",ans[i]);
			printf("\n");
			return 0;
		}
		else if(ans[i]>ans2[i])
		{
			for(int i=0;i<n;i++)
				printf("%d ",ans2[i]);
			printf("\n");
			return 0;
		}
	}
	for(int i=0;i<n;i++)
		printf("%d ",ans[i]);
	printf("\n");
	return 0;
}
### Prompt

In CPP, your task is to solve 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<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
#include<queue>
#include<bitset>
#include<complex>
using namespace std;
template<typename __T>
inline void read(__T &x)
{
    x=0;
    int f=1;char c=getchar();
    while(!isdigit(c)){if(c=='-')	f=-1;c=getchar();}
    while(isdigit(c))	{x=x*10+c-'0';c=getchar();}
    x*=f;
}
int box[100005],las[200005],edv[200005],cnt=0;
void adde(int u,int v)
{
	las[++cnt]=box[u];
	box[u]=cnt;
	edv[cnt]=v;
}
int dep[100005],dfsin[100005],dfn=0;
int tot=0;
int pre[100005];
void dfs(int now,int fa)
{
	pre[now]=fa;
	dfsin[now]=++dfn;
	for(int i=box[now];i;i=las[i])
		if(edv[i]!=fa)
		{
			dep[edv[i]]=dep[now]+1;
			dfs(edv[i],now);
		}
}
int S,T;
int n;
int ans[100005];
int ans2[100005];
int deg[100005];
bool fg=0;
bool onl[100005];
void solve(int now,int fa)
{
	dfsin[now]=++dfn;
	if(!onl[now])
	{
		if(deg[now]>1)	fg=1;
		return;
	}
	if(now==T)
	{
		ans[tot++]=dfsin[now];
		return;
	}
	int s=0;
	for(int i=box[now];i;i=las[i])
		if(edv[i]!=fa)
		{
			int v=edv[i];
			if(!onl[v])
			{
				solve(v,now);
				ans[tot++]=dfsin[v];
			}
			else
				s=v;
		}
	ans[tot++]=dfsin[now];
	solve(s,now);
}
int main()
{
	read(n);
	int a,b;
	for(int i=1;i<n;i++)
	{
		read(a);
		read(b);
		deg[a]++;
		deg[b]++;
		adde(a,b);
		adde(b,a);
	}
	dfs(1,0);
	for(int i=1;i<=n;i++)
		if(dep[i]>dep[S])	S=i;
	dfn=0;
	dfs(S,0);
	for(int i=1;i<=n;i++)
		if(dep[i]>dep[T])	T=i;
	for(int i=T;i;i=pre[i])	onl[i]=1;
	dfn=tot=0;
	solve(S,0);
	if(fg)
	{
		printf("-1\n");
		return 0;
	}
	for(int i=0;i<n;i++)
		ans2[i]=ans[i];
	swap(S,T);
	tot=dfn=0;
	solve(S,0);
	for(int i=0;i<n;i++)
	{
		if(ans[i]<ans2[i])
		{
			for(int i=0;i<n;i++)
				printf("%d ",ans[i]);
			printf("\n");
			return 0;
		}
		else if(ans[i]>ans2[i])
		{
			for(int i=0;i<n;i++)
				printf("%d ",ans2[i]);
			printf("\n");
			return 0;
		}
	}
	for(int i=0;i<n;i++)
		printf("%d ",ans[i]);
	printf("\n");
	return 0;
}
```