#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <vector>
using namespace std;
const int N=100010;
int tp,tt;
bool vis[N];
vector <int> q[N];
int v[N],head[N],to[N*2],nxt[N*2],in[N],a[N],b[N];
inline int gi() {
    int x=0,o=1;
    char ch=getchar();
    while(ch!='-'&&(ch<'0'||ch>'9')) ch=getchar();
    if(ch=='-') o=-1,ch=getchar();
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*o;
}
inline int dfs(int x,int fa) {
    if(vis[x]) return 0;
    vis[x]=1,v[++tp]=q[x].size();
    for(int i=head[x];i;i=nxt[i])
	if(to[i]!=fa) return dfs(to[i],x);
    return x;
}
int main() {
    int n;
    cin>>n;
    for(int i=1;i<n;i++)
	++in[a[i]=gi()],++in[b[i]=gi()];
    for(int i=1;i<n;i++) {
	int x=a[i],y=b[i];
	if(in[x]>1&&in[y]>1) {
	    to[++tt]=y,nxt[tt]=head[x],head[x]=tt;
	    to[++tt]=x,nxt[tt]=head[y],head[y]=tt;
	}
	if(in[x]==1) q[y].push_back(x);
	if(in[y]==1) q[x].push_back(y);
    }
    if(n==2) return puts("1\n2"),0;
    for(int i=1;i<=n;i++)
	if(in[i]>1) {
	    int x=dfs(i,0);
	    if(!x) return puts("-1"),0;
	    memset(vis,0,sizeof(vis));
	    tp=0,x=dfs(x,0);
	    for(int j=1;j<=n;j++)
		if(in[j]>1&&!vis[j]) x=0;
	    if(!x) return puts("-1"),0;
	    break;
	}
    int l=1,r=tp,t=0;
    while(l<=r) {
	if(v[l]<v[r]) break;
	if(v[l]>v[r]) {t=1;break;}
	++l,--r;
    }
    if(t) reverse(v+1,v+1+tp);
    --v[1],--v[tp];
    ++tp,t=2;
    cout<<1<<' ';
    for(int i=1;i<=tp;i++) {
	int x=v[i];
	for(int j=1;j<=x;j++) printf("%d ",t+j);
	printf("%d ",t),t+=x+1;
    }
    return 0;
}
### Prompt

Please formulate 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 <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <vector>
using namespace std;
const int N=100010;
int tp,tt;
bool vis[N];
vector <int> q[N];
int v[N],head[N],to[N*2],nxt[N*2],in[N],a[N],b[N];
inline int gi() {
    int x=0,o=1;
    char ch=getchar();
    while(ch!='-'&&(ch<'0'||ch>'9')) ch=getchar();
    if(ch=='-') o=-1,ch=getchar();
    while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
    return x*o;
}
inline int dfs(int x,int fa) {
    if(vis[x]) return 0;
    vis[x]=1,v[++tp]=q[x].size();
    for(int i=head[x];i;i=nxt[i])
	if(to[i]!=fa) return dfs(to[i],x);
    return x;
}
int main() {
    int n;
    cin>>n;
    for(int i=1;i<n;i++)
	++in[a[i]=gi()],++in[b[i]=gi()];
    for(int i=1;i<n;i++) {
	int x=a[i],y=b[i];
	if(in[x]>1&&in[y]>1) {
	    to[++tt]=y,nxt[tt]=head[x],head[x]=tt;
	    to[++tt]=x,nxt[tt]=head[y],head[y]=tt;
	}
	if(in[x]==1) q[y].push_back(x);
	if(in[y]==1) q[x].push_back(y);
    }
    if(n==2) return puts("1\n2"),0;
    for(int i=1;i<=n;i++)
	if(in[i]>1) {
	    int x=dfs(i,0);
	    if(!x) return puts("-1"),0;
	    memset(vis,0,sizeof(vis));
	    tp=0,x=dfs(x,0);
	    for(int j=1;j<=n;j++)
		if(in[j]>1&&!vis[j]) x=0;
	    if(!x) return puts("-1"),0;
	    break;
	}
    int l=1,r=tp,t=0;
    while(l<=r) {
	if(v[l]<v[r]) break;
	if(v[l]>v[r]) {t=1;break;}
	++l,--r;
    }
    if(t) reverse(v+1,v+1+tp);
    --v[1],--v[tp];
    ++tp,t=2;
    cout<<1<<' ';
    for(int i=1;i<=tp;i++) {
	int x=v[i];
	for(int j=1;j<=x;j++) printf("%d ",t+j);
	printf("%d ",t),t+=x+1;
    }
    return 0;
}
```