#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=100005,p=1000000007;
int read(){
	int f=1,g=0;
	char ch=getchar();
	for (;!isdigit(ch);ch=getchar()) if (ch=='-') f=-1;
	for (;isdigit(ch);ch=getchar()) g=g*10+ch-'0';
	return f*g;
}
int n,m,l,r,b[N],deg[N],siz[N],ans1[N],ans2[N];
pair<int,int> a[N];
vector<int> e[N];
void dfs(int x,int fa){
	if (e[x].size()>2){puts("-1");exit(0);}
	b[++m]=x;
	for (auto y : e[x])
	if (y!=fa) dfs(y,x);
}
void getans(int *ans){
	int cnt=0;
	for (int i=l;i<=r;i++){
		int t=b[i]+1;
		ans[cnt+t]=cnt+1;
		for (int j=1;j<=b[i];j++)
		ans[cnt+j]=cnt+1+j;
		cnt+=t;
	}
}
int main(){
	n=read();
	for (int i=1;i<n;i++){
		int u=read(),v=read();
		a[i]=make_pair(u,v);
		deg[u]++;deg[v]++;
	}
	for (int i=1;i<=n;i++) deg[i]--;
	for (int i=1;i<n;i++){
		int u=a[i].first,v=a[i].second;
		if (deg[u]&&deg[v]){e[u].push_back(v);e[v].push_back(u);}
		else if (deg[u]) siz[u]++;
		else siz[v]++;
	}
	for (int i=1;i<=n;i++)
	if (deg[i]&&(e[i].size()<=1)) {dfs(i,0);break;}
	for (int i=1;i<=m;i++) b[i]=siz[b[i]];
	l=1;r=m;
	if (b[l]){b[l]--;l--;}
	if (b[r]){b[r]--;r++;}
	getans(ans1);
	reverse(b+l,b+r+1);
	getans(ans2);
	int fl=0;
	for (int i=1;i<=n;i++)
	if (ans1[i]!=ans2[i]) {fl=(ans1[i]<ans2[i]);break;}
	for (int i=1;i<=n;i++) printf("%d ",(fl) ? ans1[i] : ans2[i]);
	printf("\n");
	return 0;
}
### Prompt

Create a solution in cpp for 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;
typedef long long ll;
const int N=100005,p=1000000007;
int read(){
	int f=1,g=0;
	char ch=getchar();
	for (;!isdigit(ch);ch=getchar()) if (ch=='-') f=-1;
	for (;isdigit(ch);ch=getchar()) g=g*10+ch-'0';
	return f*g;
}
int n,m,l,r,b[N],deg[N],siz[N],ans1[N],ans2[N];
pair<int,int> a[N];
vector<int> e[N];
void dfs(int x,int fa){
	if (e[x].size()>2){puts("-1");exit(0);}
	b[++m]=x;
	for (auto y : e[x])
	if (y!=fa) dfs(y,x);
}
void getans(int *ans){
	int cnt=0;
	for (int i=l;i<=r;i++){
		int t=b[i]+1;
		ans[cnt+t]=cnt+1;
		for (int j=1;j<=b[i];j++)
		ans[cnt+j]=cnt+1+j;
		cnt+=t;
	}
}
int main(){
	n=read();
	for (int i=1;i<n;i++){
		int u=read(),v=read();
		a[i]=make_pair(u,v);
		deg[u]++;deg[v]++;
	}
	for (int i=1;i<=n;i++) deg[i]--;
	for (int i=1;i<n;i++){
		int u=a[i].first,v=a[i].second;
		if (deg[u]&&deg[v]){e[u].push_back(v);e[v].push_back(u);}
		else if (deg[u]) siz[u]++;
		else siz[v]++;
	}
	for (int i=1;i<=n;i++)
	if (deg[i]&&(e[i].size()<=1)) {dfs(i,0);break;}
	for (int i=1;i<=m;i++) b[i]=siz[b[i]];
	l=1;r=m;
	if (b[l]){b[l]--;l--;}
	if (b[r]){b[r]--;r++;}
	getans(ans1);
	reverse(b+l,b+r+1);
	getans(ans2);
	int fl=0;
	for (int i=1;i<=n;i++)
	if (ans1[i]!=ans2[i]) {fl=(ans1[i]<ans2[i]);break;}
	for (int i=1;i<=n;i++) printf("%d ",(fl) ? ans1[i] : ans2[i]);
	printf("\n");
	return 0;
}
```