#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<vector>
#include<set>
#include<map>
#include<string>
#define fi first
#define se second
#define L(i,u) for (register int i=head[u]; i; i=nxt[i])
#define rep(i,a,b) for (register int i=a; i<=b; i++)
#define per(i,a,b) for (register int i=a; i>=b; i--)
using namespace std;
typedef long long ll;
typedef pair<int,int> Pii;
typedef vector<int> Vi;
template<class T> inline void read(T &x) {
	x=0; char c=getchar(); int f=1;
	while (c<'0'||c>'9') {if (c=='-') f=-1; c=getchar();}
	while (c>='0'&&c<='9') {x=10*x+c-'0'; c=getchar();} x*=f;
}
const int N = 555555,inf=0x3f3f3f3f;
int n,head[N],to[N],nxt[N],edgenum,dis[N],q[N],pre[N],deg[N];
void add(int u, int v){
	to[++edgenum]=v;nxt[edgenum]=head[u];head[u]=edgenum;
}
int bfs(int s){
	memset(dis,inf,sizeof(dis));dis[s]=0;
	int f=0,r=1;q[f]=s;int res=s;
	while(f!=r){
		int u=q[f++];
		L(i,u)if(dis[to[i]]>dis[u]+1){
			dis[to[i]]=dis[u]+1;q[r++]=to[i];pre[to[i]]=u;
			if(dis[to[i]]>dis[res])res=to[i];
		}
	}
	return res;
}
int a[N],b[N],c[N],len,mrk[N];
void solve(int *a){
//	rep(i,1,len)printf("[%d]\n",a[i]);
	int num=0,sz=0;rep(i,1,len){
		int x=++num;rep(j,1,a[i])c[++sz]=++num;c[++sz]=x;
	}
	rep(i,1,n)printf("%d ",c[i]);
}
int main() {
	read(n);rep(i,1,n-1){int u,v;read(u);read(v);add(u,v);add(v,u);deg[u]++;deg[v]++;}
	int S=bfs(1),T=bfs(S);
	for(int u=T;;u=pre[u]){mrk[u]=1;if(u==S)break;}
	for(int u=T;;u=pre[u]){
		len++;L(i,u)if(!mrk[to[i]]){
			a[len]++;if(deg[to[i]]>1){puts("-1");return 0;}
		}if(u==S)break;
	}
	rep(i,1,len)b[i]=a[len-i+1];
	rep(i,1,len)if(a[i]!=b[i]){
		if(a[i]<b[i])solve(a);else solve(b);return 0;
	}
	solve(a);
	return 0;
}
### 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<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<vector>
#include<set>
#include<map>
#include<string>
#define fi first
#define se second
#define L(i,u) for (register int i=head[u]; i; i=nxt[i])
#define rep(i,a,b) for (register int i=a; i<=b; i++)
#define per(i,a,b) for (register int i=a; i>=b; i--)
using namespace std;
typedef long long ll;
typedef pair<int,int> Pii;
typedef vector<int> Vi;
template<class T> inline void read(T &x) {
	x=0; char c=getchar(); int f=1;
	while (c<'0'||c>'9') {if (c=='-') f=-1; c=getchar();}
	while (c>='0'&&c<='9') {x=10*x+c-'0'; c=getchar();} x*=f;
}
const int N = 555555,inf=0x3f3f3f3f;
int n,head[N],to[N],nxt[N],edgenum,dis[N],q[N],pre[N],deg[N];
void add(int u, int v){
	to[++edgenum]=v;nxt[edgenum]=head[u];head[u]=edgenum;
}
int bfs(int s){
	memset(dis,inf,sizeof(dis));dis[s]=0;
	int f=0,r=1;q[f]=s;int res=s;
	while(f!=r){
		int u=q[f++];
		L(i,u)if(dis[to[i]]>dis[u]+1){
			dis[to[i]]=dis[u]+1;q[r++]=to[i];pre[to[i]]=u;
			if(dis[to[i]]>dis[res])res=to[i];
		}
	}
	return res;
}
int a[N],b[N],c[N],len,mrk[N];
void solve(int *a){
//	rep(i,1,len)printf("[%d]\n",a[i]);
	int num=0,sz=0;rep(i,1,len){
		int x=++num;rep(j,1,a[i])c[++sz]=++num;c[++sz]=x;
	}
	rep(i,1,n)printf("%d ",c[i]);
}
int main() {
	read(n);rep(i,1,n-1){int u,v;read(u);read(v);add(u,v);add(v,u);deg[u]++;deg[v]++;}
	int S=bfs(1),T=bfs(S);
	for(int u=T;;u=pre[u]){mrk[u]=1;if(u==S)break;}
	for(int u=T;;u=pre[u]){
		len++;L(i,u)if(!mrk[to[i]]){
			a[len]++;if(deg[to[i]]>1){puts("-1");return 0;}
		}if(u==S)break;
	}
	rep(i,1,len)b[i]=a[len-i+1];
	rep(i,1,len)if(a[i]!=b[i]){
		if(a[i]<b[i])solve(a);else solve(b);return 0;
	}
	solve(a);
	return 0;
}
```