#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
#include<queue>
#include<cmath>
#include<cstdlib>
#define LL long long
#define LD long double
using namespace std;
const int NN=100000 +117;
const int MM=100000 +117;
int read(){
	int fl=1,x;char c;
	for(c=getchar();(c<'0'||c>'9')&&c!='-';c=getchar());
	if(c=='-'){fl=-1;c=getchar();}
	for(x=0;c>='0'&&c<='9';c=getchar())
		x=(x<<3)+(x<<1)+c-'0';
	return x*fl;
}
void open(){
	freopen("a.in","r",stdin);
	//freopen("a.out","w",stdout);
}
void close(){
	fclose(stdin);
	fclose(stdout);
}


int m,n;
int deg[NN]={};
vector<int> p[NN];
int l;
int r;
int acnt=0;
int a[NN]={};
void dfs(int x,int fa){
	int xt=0;
	int cnt=0;
	for(int i=0;i<p[x].size();++i){
		int cur=p[x][i];
		if(deg[cur]!=1){
			if(cur!=fa)xt=cur;
		}
		else ++cnt;
	}
	a[++acnt]=cnt;
	if(xt)dfs(xt,x);
}
int res[2][NN]={};
void getres(int k){
	int cc=0;
	int now=0;
	res[k][++now]=++cc;
	for(int i=1;i<=acnt;++i){
		int s=a[i];
		for(int j=1;j<=s;++j){
			res[k][++now]=cc+1+j;
		}
		res[k][++now]=cc+1;
		cc+=s+1;
	}
	res[k][++now]=++cc;
}
bool cmp(){
	int pos=1;
	while(pos<=n&&res[0][pos]==res[1][pos])++pos;
	return res[1][pos]<res[0][pos];
}
int main(){
	//open();
	n=read();
	if(n==2){
		printf("1 2\n");
		return 0;
	}
	for(int i=1;i<n;++i){
		int x=read(),y=read();
		deg[x]++,deg[y]++;
		p[x].push_back(y);
		p[y].push_back(x);
	}
	int duan;
	bool fl=1;
	for(int i=1;i<=n;++i){
		int cnt=0;
		for(int j=0;j<p[i].size();++j){
			int cur=p[i][j];
			if(deg[cur]!=1)++cnt;
		}
		if(cnt<=1&&deg[i]!=1)duan=i;
		if(cnt>2)fl=0;
	}
	if(!fl){
		printf("-1\n");
		return 0;
	}
	dfs(duan,0);
	a[1]--;
	a[acnt]--;
	getres(1);
	for(int i=1;i<=acnt/2;++i)swap(a[i],a[acnt+1-i]);
	getres(0);
	int k=cmp();
	for(int i=1;i<=n;++i){
		printf("%d ",res[k][i]);
	}
	putchar('\n');
	close();
	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<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
#include<queue>
#include<cmath>
#include<cstdlib>
#define LL long long
#define LD long double
using namespace std;
const int NN=100000 +117;
const int MM=100000 +117;
int read(){
	int fl=1,x;char c;
	for(c=getchar();(c<'0'||c>'9')&&c!='-';c=getchar());
	if(c=='-'){fl=-1;c=getchar();}
	for(x=0;c>='0'&&c<='9';c=getchar())
		x=(x<<3)+(x<<1)+c-'0';
	return x*fl;
}
void open(){
	freopen("a.in","r",stdin);
	//freopen("a.out","w",stdout);
}
void close(){
	fclose(stdin);
	fclose(stdout);
}


int m,n;
int deg[NN]={};
vector<int> p[NN];
int l;
int r;
int acnt=0;
int a[NN]={};
void dfs(int x,int fa){
	int xt=0;
	int cnt=0;
	for(int i=0;i<p[x].size();++i){
		int cur=p[x][i];
		if(deg[cur]!=1){
			if(cur!=fa)xt=cur;
		}
		else ++cnt;
	}
	a[++acnt]=cnt;
	if(xt)dfs(xt,x);
}
int res[2][NN]={};
void getres(int k){
	int cc=0;
	int now=0;
	res[k][++now]=++cc;
	for(int i=1;i<=acnt;++i){
		int s=a[i];
		for(int j=1;j<=s;++j){
			res[k][++now]=cc+1+j;
		}
		res[k][++now]=cc+1;
		cc+=s+1;
	}
	res[k][++now]=++cc;
}
bool cmp(){
	int pos=1;
	while(pos<=n&&res[0][pos]==res[1][pos])++pos;
	return res[1][pos]<res[0][pos];
}
int main(){
	//open();
	n=read();
	if(n==2){
		printf("1 2\n");
		return 0;
	}
	for(int i=1;i<n;++i){
		int x=read(),y=read();
		deg[x]++,deg[y]++;
		p[x].push_back(y);
		p[y].push_back(x);
	}
	int duan;
	bool fl=1;
	for(int i=1;i<=n;++i){
		int cnt=0;
		for(int j=0;j<p[i].size();++j){
			int cur=p[i][j];
			if(deg[cur]!=1)++cnt;
		}
		if(cnt<=1&&deg[i]!=1)duan=i;
		if(cnt>2)fl=0;
	}
	if(!fl){
		printf("-1\n");
		return 0;
	}
	dfs(duan,0);
	a[1]--;
	a[acnt]--;
	getres(1);
	for(int i=1;i<=acnt/2;++i)swap(a[i],a[acnt+1-i]);
	getres(0);
	int k=cmp();
	for(int i=1;i<=n;++i){
		printf("%d ",res[k][i]);
	}
	putchar('\n');
	close();
	return 0;
}
```