#include<bits/stdc++.h>
using namespace std;
const int N=2e5+10;
int w[N],ne[N],la[N],t,n,u,v,dep[N];
void link(int x,int y){
	w[++t]=y;
	ne[t]=la[x];
	la[x]=t;
}
int dfs(int x,int fa=0){
	int res=x;
	for (int y=la[x];y;y=ne[y]){
		int z=w[y];
		if (z==fa)continue;
		dep[z]=dep[x]+1;
		int z1=dfs(z,x);
		if (dep[z1]>dep[res])res=z1;
	}return res;
}
int a1[N],cnt[N],tag[N];
int go(int x,int tar,int fa){
	if (x==tar){
		tag[x]=1;
		a1[dep[x]]=x;
		return 1;
	}
	for (int y=la[x];y;y=ne[y]){
		int z=w[y];
		if (z==fa)continue;
		dep[z]=dep[x]+1;
		if (go(z,tar,x)){
			tag[x]=1;
			a1[dep[x]]=x;
			return 1;
		}
	}return 0;
}
int a[N],b[N],t1;
int main(){
	cin>>n;
	for (int i=1;i<n;i++){
		int x,y;
		scanf("%d%d",&x,&y);
		link(x,y);
		link(y,x);
	}
	int u=dfs(1),v=dfs(u);
	dep[u]=0;
	go(u,v,0);
	for (int x=1;x<=n;x++){
		if (tag[x])continue;
		int cnt1=0;
		for (int y=la[x];y;y=ne[y]){
			int z=w[y];
			if (tag[z])
				cnt1++,cnt[dep[z]]++;
		}
		if (cnt1!=1){
			cout<<-1<<endl;
			return 0;
		}
	}
	t=0;
	for (int i=0;i<=dep[v];i++){
		a[++t]=t;
		for (int j=1;j<=cnt[i];j++)
			a[++t]=t,swap(a[t],a[t-1]);
	}
	t=0;
	reverse(cnt,cnt+dep[v]+1);
	for (int i=0;i<=dep[v];i++){
		b[++t]=t;
		for (int j=1;j<=cnt[i];j++)
			b[++t]=t,swap(b[t],b[t-1]);
	}
	for (int i=1;i<=n;i++){
		if (a[i]<b[i])break;
		if (a[i]>b[i]){
			memcpy(a,b,sizeof a);
			break;
		}
	}
	for (int i=1;i<=n;i++)printf("%d ",a[i]);
}
### 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<bits/stdc++.h>
using namespace std;
const int N=2e5+10;
int w[N],ne[N],la[N],t,n,u,v,dep[N];
void link(int x,int y){
	w[++t]=y;
	ne[t]=la[x];
	la[x]=t;
}
int dfs(int x,int fa=0){
	int res=x;
	for (int y=la[x];y;y=ne[y]){
		int z=w[y];
		if (z==fa)continue;
		dep[z]=dep[x]+1;
		int z1=dfs(z,x);
		if (dep[z1]>dep[res])res=z1;
	}return res;
}
int a1[N],cnt[N],tag[N];
int go(int x,int tar,int fa){
	if (x==tar){
		tag[x]=1;
		a1[dep[x]]=x;
		return 1;
	}
	for (int y=la[x];y;y=ne[y]){
		int z=w[y];
		if (z==fa)continue;
		dep[z]=dep[x]+1;
		if (go(z,tar,x)){
			tag[x]=1;
			a1[dep[x]]=x;
			return 1;
		}
	}return 0;
}
int a[N],b[N],t1;
int main(){
	cin>>n;
	for (int i=1;i<n;i++){
		int x,y;
		scanf("%d%d",&x,&y);
		link(x,y);
		link(y,x);
	}
	int u=dfs(1),v=dfs(u);
	dep[u]=0;
	go(u,v,0);
	for (int x=1;x<=n;x++){
		if (tag[x])continue;
		int cnt1=0;
		for (int y=la[x];y;y=ne[y]){
			int z=w[y];
			if (tag[z])
				cnt1++,cnt[dep[z]]++;
		}
		if (cnt1!=1){
			cout<<-1<<endl;
			return 0;
		}
	}
	t=0;
	for (int i=0;i<=dep[v];i++){
		a[++t]=t;
		for (int j=1;j<=cnt[i];j++)
			a[++t]=t,swap(a[t],a[t-1]);
	}
	t=0;
	reverse(cnt,cnt+dep[v]+1);
	for (int i=0;i<=dep[v];i++){
		b[++t]=t;
		for (int j=1;j<=cnt[i];j++)
			b[++t]=t,swap(b[t],b[t-1]);
	}
	for (int i=1;i<=n;i++){
		if (a[i]<b[i])break;
		if (a[i]>b[i]){
			memcpy(a,b,sizeof a);
			break;
		}
	}
	for (int i=1;i<=n;i++)printf("%d ",a[i]);
}
```