#include<bits/stdc++.h>

using namespace std;

#define N 800000

int n,x,y,tot,S,T,xb,cnt,head[N],q[N],mx[N],num[N],a[N];
bool vis[N];
struct edge{int v,nxt;}e[N];

void add(int x,int y){e[++tot].v=y; e[tot].nxt=head[x]; head[x]=tot;}

void bfs(){
	memset(vis,0,sizeof vis);
	int tt=0,ww=0; q[++ww]=S; vis[S]=1;
	while (tt<ww){
		int u=q[++tt];
		for (int i=head[u],v;i;i=e[i].nxt)
			if (!vis[v=e[i].v]){
				vis[v]=1;
				q[++ww]=v;
			}
	}
	T=q[ww];
}

void dfs(int u,int fa){
	int tmp=0; mx[u]=0;
	for (int i=head[u],v;i;i=e[i].nxt)
		if ((v=e[i].v)!=fa){
			dfs(v,u); mx[u]=max(mx[u],mx[v]);
			if (mx[v]>1) ++tmp;
			else if (v!=T) ++num[u];
		}
	if (tmp>1){puts("-1"); exit(0);}
	++mx[u];
	if (u!=S&&u!=T&&mx[u]>1) a[++xb]=num[u];
}

void print(int u,int fa){
	int t=++cnt;
	for (int i=head[u],v;i;i=e[i].nxt)
		if ((v=e[i].v)!=fa&&mx[v]==1&&v!=T)
			printf("%d ",++cnt);
	printf("%d ",t);
	for (int i=head[u],v;i;i=e[i].nxt)
		if ((v=e[i].v)!=fa&&(mx[v]>1||v==T))
			print(v,u);
}

int main(){
	scanf("%d",&n);
	for (int i=1;i<n;++i){
		scanf("%d%d",&x,&y);
		add(x,y); add(y,x);
	}
	S=1; bfs(); S=T; bfs();
	dfs(S,0); reverse(a+1,a+xb+1);
	bool fl=1;
	for (int i=1;i<=xb;++i){
		if (a[i]<a[xb-i+1]){fl=1; break;}
		if (a[i]>a[xb-i+1]){fl=0; break;}
	}
	xb=0;
	if (fl) print(S,0);
	else{swap(S,T); dfs(S,0); print(S,0);}
	
	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<bits/stdc++.h>

using namespace std;

#define N 800000

int n,x,y,tot,S,T,xb,cnt,head[N],q[N],mx[N],num[N],a[N];
bool vis[N];
struct edge{int v,nxt;}e[N];

void add(int x,int y){e[++tot].v=y; e[tot].nxt=head[x]; head[x]=tot;}

void bfs(){
	memset(vis,0,sizeof vis);
	int tt=0,ww=0; q[++ww]=S; vis[S]=1;
	while (tt<ww){
		int u=q[++tt];
		for (int i=head[u],v;i;i=e[i].nxt)
			if (!vis[v=e[i].v]){
				vis[v]=1;
				q[++ww]=v;
			}
	}
	T=q[ww];
}

void dfs(int u,int fa){
	int tmp=0; mx[u]=0;
	for (int i=head[u],v;i;i=e[i].nxt)
		if ((v=e[i].v)!=fa){
			dfs(v,u); mx[u]=max(mx[u],mx[v]);
			if (mx[v]>1) ++tmp;
			else if (v!=T) ++num[u];
		}
	if (tmp>1){puts("-1"); exit(0);}
	++mx[u];
	if (u!=S&&u!=T&&mx[u]>1) a[++xb]=num[u];
}

void print(int u,int fa){
	int t=++cnt;
	for (int i=head[u],v;i;i=e[i].nxt)
		if ((v=e[i].v)!=fa&&mx[v]==1&&v!=T)
			printf("%d ",++cnt);
	printf("%d ",t);
	for (int i=head[u],v;i;i=e[i].nxt)
		if ((v=e[i].v)!=fa&&(mx[v]>1||v==T))
			print(v,u);
}

int main(){
	scanf("%d",&n);
	for (int i=1;i<n;++i){
		scanf("%d%d",&x,&y);
		add(x,y); add(y,x);
	}
	S=1; bfs(); S=T; bfs();
	dfs(S,0); reverse(a+1,a+xb+1);
	bool fl=1;
	for (int i=1;i<=xb;++i){
		if (a[i]<a[xb-i+1]){fl=1; break;}
		if (a[i]>a[xb-i+1]){fl=0; break;}
	}
	xb=0;
	if (fl) print(S,0);
	else{swap(S,T); dfs(S,0); print(S,0);}
	
	return 0;
}
			
```