#include<bits/stdc++.h>
using namespace std;
int read( ){
	int x=0,y=1;char ch=' ';
	for(;(ch!='-' && (ch>'9' || ch<'0'));ch=getchar( ));
	if(ch=='-')y=-1,ch=getchar( );
	for(;ch>='0' && ch<='9';ch=getchar( ))x=x*10+ch-48;
	return x*y;
}
const int N=1e5+10;
struct node{
	int to,nex;
}e[N<<1];
int bgn[N],d[N],fr[N],sz[N],b[N];
int n,cnt;
vector<int>v;
void add(int x,int y){
	e[++cnt]=(node){y,bgn[x]};bgn[x]=cnt;
}
void dfs(int x,int k){
	d[x]=k;
	for(register int i=bgn[x];i;i=e[i].nex){
		int y=e[i].to;if(d[y])continue;
		fr[y]=x;dfs(y,k+1);
	}
}
void getsz(int x,int fa){
	sz[x]=1;
	for(register int i=bgn[x];i;i=e[i].nex){
		int y=e[i].to;if(y==fa)continue;
		getsz(y,x);sz[x]+=sz[y];
	}
}
int main( ){
	int x,y,z,k;
	n=read();
	for(register int i=1;i<n;++i){
		x=read();y=read();
		add(x,y);add(y,x);
	}
	dfs(1,1);
	int mn=0;
	for(register int i=1;i<=n;++i)if(d[i]>mn){
		mn=d[i];x=i;
	}
	for(register int i=1;i<=n;++i)d[i]=fr[i]=0;
	dfs(x,1);
	mn=0;
	for(register int i=1;i<=n;++i)if(d[i]>mn){
		mn=d[i];x=i;
	}
	getsz(x,0);
	int u=x;
	while(u){
		b[u]=1;	
		u=fr[u];
	}
	u=x;
	while(u){
		for(register int i=bgn[u];i;i=e[i].nex){
			int y=e[i].to;if(b[y])continue;
			if(sz[y]>1){
				puts("-1");return 0;
			}
		}
		v.push_back(sz[u]-sz[fr[u]]);
		u=fr[u];
	}
	for(register int i=0;i<v.size();++i){
		if(v[i]<v[v.size()-i-1])break;
		if(v[i]>v[v.size()-i-1]){
			reverse(v.begin(),v.end());
			break;
		}
	}
	x=0;
	for(register int i=0;i<v.size();++i){
		for(register int j=x+2;j<=x+v[i];++j)printf("%d ",j);
		printf("%d ",x+1);
		x=x+v[i];
	}
	
	return 0;
}

### Prompt

Please create a solution in Cpp 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>
using namespace std;
int read( ){
	int x=0,y=1;char ch=' ';
	for(;(ch!='-' && (ch>'9' || ch<'0'));ch=getchar( ));
	if(ch=='-')y=-1,ch=getchar( );
	for(;ch>='0' && ch<='9';ch=getchar( ))x=x*10+ch-48;
	return x*y;
}
const int N=1e5+10;
struct node{
	int to,nex;
}e[N<<1];
int bgn[N],d[N],fr[N],sz[N],b[N];
int n,cnt;
vector<int>v;
void add(int x,int y){
	e[++cnt]=(node){y,bgn[x]};bgn[x]=cnt;
}
void dfs(int x,int k){
	d[x]=k;
	for(register int i=bgn[x];i;i=e[i].nex){
		int y=e[i].to;if(d[y])continue;
		fr[y]=x;dfs(y,k+1);
	}
}
void getsz(int x,int fa){
	sz[x]=1;
	for(register int i=bgn[x];i;i=e[i].nex){
		int y=e[i].to;if(y==fa)continue;
		getsz(y,x);sz[x]+=sz[y];
	}
}
int main( ){
	int x,y,z,k;
	n=read();
	for(register int i=1;i<n;++i){
		x=read();y=read();
		add(x,y);add(y,x);
	}
	dfs(1,1);
	int mn=0;
	for(register int i=1;i<=n;++i)if(d[i]>mn){
		mn=d[i];x=i;
	}
	for(register int i=1;i<=n;++i)d[i]=fr[i]=0;
	dfs(x,1);
	mn=0;
	for(register int i=1;i<=n;++i)if(d[i]>mn){
		mn=d[i];x=i;
	}
	getsz(x,0);
	int u=x;
	while(u){
		b[u]=1;	
		u=fr[u];
	}
	u=x;
	while(u){
		for(register int i=bgn[u];i;i=e[i].nex){
			int y=e[i].to;if(b[y])continue;
			if(sz[y]>1){
				puts("-1");return 0;
			}
		}
		v.push_back(sz[u]-sz[fr[u]]);
		u=fr[u];
	}
	for(register int i=0;i<v.size();++i){
		if(v[i]<v[v.size()-i-1])break;
		if(v[i]>v[v.size()-i-1]){
			reverse(v.begin(),v.end());
			break;
		}
	}
	x=0;
	for(register int i=0;i<v.size();++i){
		for(register int j=x+2;j<=x+v[i];++j)printf("%d ",j);
		printf("%d ",x+1);
		x=x+v[i];
	}
	
	return 0;
}

```