# include <iostream>
# include <cstdio>
using namespace std;
const int N = 1e5 + 12;
int head[N],dt,dep[N],mx,rt,n,sz[N],fa[N],num[N],que[N],tot,ans[2][N];
struct Edge{int to,nex;}edge[N <<1];bool vis[N];
void AddEdge(int u,int v){edge[++dt] = (Edge){v,head[u]};head[u] = dt;}
void dfs(int u)
{
	if(dep[u] > dep[mx])mx = u;sz[u] = 1;
	for(int i = head[u];i;i = edge[i].nex)if(edge[i].to != fa[u])
	{
		dep[edge[i].to] = dep[u] + 1;
		fa[edge[i].to] = u;
		dfs(edge[i].to);
		sz[u] += sz[edge[i].to];
	}
}
int main()
{
  int x,y;scanf("%d",&n);
  for(int i = 1;i < n;i++)scanf("%d %d",&x,&y),AddEdge(x,y),AddEdge(y,x);
  dfs(1);dep[rt = mx] = fa[rt] = 0;dfs(rt);x = mx;
  while(x)vis[x] = true,que[++tot] = x,x = fa[x];
  for(int i = 1;i <= n;i++)if(!vis[i] && sz[i] != 1){puts("-1");return 0;}
  else num[fa[i]] += !vis[i];
  for(int i = 1,p = 0,t = 0;i <= tot;i++,t++)
  {
  	ans[0][t + num[que[i]] + 1] = ++p;
  	for(int j = 1;j <= num[que[i]];j++)
  	ans[0][++t] = ++p;
  } 
  for(int i = tot,p = 0,t = 0;i >= 1;i--,t++)
  {
  	ans[1][t + num[que[i]] + 1] = ++p;
  	for(int j = 1;j <= num[que[i]];j++)
  	ans[1][++t] = ++p;
  } 
  int cur = -1;
  for(int i = 1;i <= n && cur == -1;i++)
  {
  	if(ans[0][i] < ans[1][i])cur = 0;
  	if(ans[0][i] > ans[1][i])cur = 1;
  }
  if(cur == -1)cur = 0;
  printf("%d",ans[cur][1]);
  for(int i = 2;i <= n;i++)printf(" %d",ans[cur][i]);
}
### 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 <iostream>
# include <cstdio>
using namespace std;
const int N = 1e5 + 12;
int head[N],dt,dep[N],mx,rt,n,sz[N],fa[N],num[N],que[N],tot,ans[2][N];
struct Edge{int to,nex;}edge[N <<1];bool vis[N];
void AddEdge(int u,int v){edge[++dt] = (Edge){v,head[u]};head[u] = dt;}
void dfs(int u)
{
	if(dep[u] > dep[mx])mx = u;sz[u] = 1;
	for(int i = head[u];i;i = edge[i].nex)if(edge[i].to != fa[u])
	{
		dep[edge[i].to] = dep[u] + 1;
		fa[edge[i].to] = u;
		dfs(edge[i].to);
		sz[u] += sz[edge[i].to];
	}
}
int main()
{
  int x,y;scanf("%d",&n);
  for(int i = 1;i < n;i++)scanf("%d %d",&x,&y),AddEdge(x,y),AddEdge(y,x);
  dfs(1);dep[rt = mx] = fa[rt] = 0;dfs(rt);x = mx;
  while(x)vis[x] = true,que[++tot] = x,x = fa[x];
  for(int i = 1;i <= n;i++)if(!vis[i] && sz[i] != 1){puts("-1");return 0;}
  else num[fa[i]] += !vis[i];
  for(int i = 1,p = 0,t = 0;i <= tot;i++,t++)
  {
  	ans[0][t + num[que[i]] + 1] = ++p;
  	for(int j = 1;j <= num[que[i]];j++)
  	ans[0][++t] = ++p;
  } 
  for(int i = tot,p = 0,t = 0;i >= 1;i--,t++)
  {
  	ans[1][t + num[que[i]] + 1] = ++p;
  	for(int j = 1;j <= num[que[i]];j++)
  	ans[1][++t] = ++p;
  } 
  int cur = -1;
  for(int i = 1;i <= n && cur == -1;i++)
  {
  	if(ans[0][i] < ans[1][i])cur = 0;
  	if(ans[0][i] > ans[1][i])cur = 1;
  }
  if(cur == -1)cur = 0;
  printf("%d",ans[cur][1]);
  for(int i = 2;i <= n;i++)printf(" %d",ans[cur][i]);
}
```