#include<bits/stdc++.h>
#define EB emplace_back
#define PB push_back
#define SZ(x) ((int)(x).size())
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
using namespace std;
const int N=100005;
int n,dep[N];
int pp,lnk[N],nxt[N*2],to[N*2];
void ae(int k1,int k2){to[++pp]=k2,nxt[pp]=lnk[k1],lnk[k1]=pp;}
void dfs(int k1,int k2){
	dep[k1]=dep[k2]+1;
	for(int i=lnk[k1];i;i=nxt[i])if(to[i]!=k2)dfs(to[i],k1);
}
vector<int>lu;
vector<int>G[N];
bool getlu(int k1,int k2,int k3){
	if(k1==k3){
		lu.EB(k1);
		return 1;
	}
	for(int i=lnk[k1];i;i=nxt[i])if(to[i]!=k2){
		if(getlu(to[i],k1,k3)){
			lu.EB(k1);
			return 1;
		}
	}
	return 0;
}
int main(){
	scanf("%d",&n);
	rep(i,2,n){
		int k1,k2;scanf("%d%d",&k1,&k2);
		ae(k1,k2),ae(k2,k1);
	}
	dfs(1,0);
	int rt1=max_element(dep+1,dep+1+n)-dep;
	dfs(rt1,0);
	int rt2=max_element(dep+1,dep+1+n)-dep;
	getlu(rt1,0,rt2);
	vector<bool>vis(n+1);
	vis[0]=1;
	for(auto k1:lu)vis[k1]=1;
	for(auto k1:lu){
		for(int i=lnk[k1];i;i=nxt[i]){
			if(!vis[to[i]]){
				G[k1].EB(to[i]);
				vis[to[i]]=1;
			}
		}
	}
	if(count(vis.begin(),vis.end(),0)){
		puts("-1");
		return 0;
	}
	vector<int>p1(n+1),p2(n+1);
	int ind=1;
	for(auto k1:lu){
		int cur=ind;
		rep(_,0,SZ(G[k1])-1)p1[ind]=ind+1,++ind;
		p1[ind++]=cur;
	}
	ind=1;
	reverse(lu.begin(),lu.end());
	for(auto k1:lu){
		int cur=ind;
		rep(_,0,SZ(G[k1])-1)p2[ind]=ind+1,++ind;
		p2[ind++]=cur;
	}
	if(p2<p1)p1=p2;
	rep(i,1,n)printf("%d ",p1[i]);
	puts("");
	return 0;
}
### Prompt

Your challenge is to write a cpp solution 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>
#define EB emplace_back
#define PB push_back
#define SZ(x) ((int)(x).size())
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
using namespace std;
const int N=100005;
int n,dep[N];
int pp,lnk[N],nxt[N*2],to[N*2];
void ae(int k1,int k2){to[++pp]=k2,nxt[pp]=lnk[k1],lnk[k1]=pp;}
void dfs(int k1,int k2){
	dep[k1]=dep[k2]+1;
	for(int i=lnk[k1];i;i=nxt[i])if(to[i]!=k2)dfs(to[i],k1);
}
vector<int>lu;
vector<int>G[N];
bool getlu(int k1,int k2,int k3){
	if(k1==k3){
		lu.EB(k1);
		return 1;
	}
	for(int i=lnk[k1];i;i=nxt[i])if(to[i]!=k2){
		if(getlu(to[i],k1,k3)){
			lu.EB(k1);
			return 1;
		}
	}
	return 0;
}
int main(){
	scanf("%d",&n);
	rep(i,2,n){
		int k1,k2;scanf("%d%d",&k1,&k2);
		ae(k1,k2),ae(k2,k1);
	}
	dfs(1,0);
	int rt1=max_element(dep+1,dep+1+n)-dep;
	dfs(rt1,0);
	int rt2=max_element(dep+1,dep+1+n)-dep;
	getlu(rt1,0,rt2);
	vector<bool>vis(n+1);
	vis[0]=1;
	for(auto k1:lu)vis[k1]=1;
	for(auto k1:lu){
		for(int i=lnk[k1];i;i=nxt[i]){
			if(!vis[to[i]]){
				G[k1].EB(to[i]);
				vis[to[i]]=1;
			}
		}
	}
	if(count(vis.begin(),vis.end(),0)){
		puts("-1");
		return 0;
	}
	vector<int>p1(n+1),p2(n+1);
	int ind=1;
	for(auto k1:lu){
		int cur=ind;
		rep(_,0,SZ(G[k1])-1)p1[ind]=ind+1,++ind;
		p1[ind++]=cur;
	}
	ind=1;
	reverse(lu.begin(),lu.end());
	for(auto k1:lu){
		int cur=ind;
		rep(_,0,SZ(G[k1])-1)p2[ind]=ind+1,++ind;
		p2[ind++]=cur;
	}
	if(p2<p1)p1=p2;
	rep(i,1,n)printf("%d ",p1[i]);
	puts("");
	return 0;
}
```