#include <bits/stdc++.h>

using namespace std;

vector <int> e[100005];
int dep[100005],fa[100005];

int dfs1(int x) {
  int s=x;
  for(int i=0;i<e[x].size();i++)
    if (e[x][i]!=fa[x]) {
    	int u=e[x][i];
    	fa[u]=x;dep[u]=dep[x]+1;
    	int t=dfs1(u);
    	if (dep[t]>dep[s]) s=t;
	}
  return s;
}

int d[100005],num[100005];
bool vis[100005];

int p[100005],q[100005];

int main() {
  int n;
  scanf("%d",&n);
  for(int i=1;i<n;i++) {
  	int x,y;
  	scanf("%d%d",&x,&y);
  	e[x].push_back(y);
  	e[y].push_back(x);
  }
  int rt1=dfs1(1);
  fa[rt1]=dep[rt1]=0;
  int rt2=dfs1(rt1),cnt=0;
  for(int i=rt2;i;i=fa[i]) {
    num[++cnt]=i;
    vis[i]=1;
  }
  for(int i=1;i<=n;i++)
    if (!vis[i]) {
    	bool ok=0;
    	for(int j=0;j<e[i].size();j++)
    	  if (vis[e[i][j]]) {
    	  	d[e[i][j]]++;
    	  	ok=1;
    	  	break;
		  }
		if (!ok) {
			puts("-1");
			return 0;
		}
	}
  int s=0;
  for(int i=1;i<=cnt;i++) {
  	int x=num[i];
  	for(int j=1;j<=d[x];j++) p[s+j]=s+j+1;
  	p[s+d[x]+1]=s+1;
  	s+=d[x]+1;
  }
  s=0;
  for(int i=cnt;i>0;i--) {
  	int x=num[i];
  	for(int j=1;j<=d[x];j++) q[s+j]=s+j+1;
  	q[s+d[x]+1]=s+1;
  	s+=d[x]+1;
  }
  bool ok=0;
  for(int i=1;i<=n;i++)
    if (p[i]!=q[i]) {
    	ok=(p[i]<q[i]);
    	break;
	}
  if (ok) {
  	for(int i=1;i<=n;i++) printf("%d ",p[i]);
  	printf("\n");
  }
  else {
  	for(int i=1;i<=n;i++) printf("%d ",q[i]);
  	printf("\n");
  }
  return 0;
}
### Prompt

Generate 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>

using namespace std;

vector <int> e[100005];
int dep[100005],fa[100005];

int dfs1(int x) {
  int s=x;
  for(int i=0;i<e[x].size();i++)
    if (e[x][i]!=fa[x]) {
    	int u=e[x][i];
    	fa[u]=x;dep[u]=dep[x]+1;
    	int t=dfs1(u);
    	if (dep[t]>dep[s]) s=t;
	}
  return s;
}

int d[100005],num[100005];
bool vis[100005];

int p[100005],q[100005];

int main() {
  int n;
  scanf("%d",&n);
  for(int i=1;i<n;i++) {
  	int x,y;
  	scanf("%d%d",&x,&y);
  	e[x].push_back(y);
  	e[y].push_back(x);
  }
  int rt1=dfs1(1);
  fa[rt1]=dep[rt1]=0;
  int rt2=dfs1(rt1),cnt=0;
  for(int i=rt2;i;i=fa[i]) {
    num[++cnt]=i;
    vis[i]=1;
  }
  for(int i=1;i<=n;i++)
    if (!vis[i]) {
    	bool ok=0;
    	for(int j=0;j<e[i].size();j++)
    	  if (vis[e[i][j]]) {
    	  	d[e[i][j]]++;
    	  	ok=1;
    	  	break;
		  }
		if (!ok) {
			puts("-1");
			return 0;
		}
	}
  int s=0;
  for(int i=1;i<=cnt;i++) {
  	int x=num[i];
  	for(int j=1;j<=d[x];j++) p[s+j]=s+j+1;
  	p[s+d[x]+1]=s+1;
  	s+=d[x]+1;
  }
  s=0;
  for(int i=cnt;i>0;i--) {
  	int x=num[i];
  	for(int j=1;j<=d[x];j++) q[s+j]=s+j+1;
  	q[s+d[x]+1]=s+1;
  	s+=d[x]+1;
  }
  bool ok=0;
  for(int i=1;i<=n;i++)
    if (p[i]!=q[i]) {
    	ok=(p[i]<q[i]);
    	break;
	}
  if (ok) {
  	for(int i=1;i<=n;i++) printf("%d ",p[i]);
  	printf("\n");
  }
  else {
  	for(int i=1;i<=n;i++) printf("%d ",q[i]);
  	printf("\n");
  }
  return 0;
}
```