#include<bits/stdc++.h>
#define N 100005
#define M 200005
using namespace std;
int i,j,k,l,s,n,m,last[N],to[M],Next[M],a[N],b[N],bb,fa[N],vis[N],Ans[N],ans,ans1,x,y;
int q[N],tot,X,Y;
inline void add(int x,int y) {
	Next[++tot]=last[x]; last[x]=tot; to[tot]=y;
}
inline void dfs(int x) {
	vis[x]=1;
	s++;
	if (s>ans) ans=s,ans1=x;
	for (int i=last[x];i;i=Next[i]) if (!vis[to[i]]) dfs(to[i]);
	vis[x]=0,s--;
}
inline void build(int x) {
	int l=0,r=1; q[1]=x;
	while (l<r) {
		int k=q[++l];
		for (int i=last[k];i;i=Next[i]) if (fa[k]!=to[i]) fa[q[++r]=to[i]]=k;
	}
}
int main() {
	scanf("%d",&n);
	for (i=1;i<n;i++) scanf("%d%d",&x,&y),add(x,y),add(y,x);
	dfs(1);
	ans=0;
	X=ans1;
	dfs(ans1);
	Y=ans1;
	build(X);
	for (i=Y;i;i=fa[i]) vis[i]=1;
	for (i=1;i<=n;i++) if (!vis[i]&&!vis[fa[i]]) return puts("-1"),0;
	tot=0;
	for (i=Y;i;i=fa[i]) {
		tot++;
		for (j=last[i];j;j=Next[j]) {
			if (to[j]==fa[i]) continue;
			if (!vis[to[j]]) a[tot]++;
		}
	}
	for (i=1;i<=tot;i++) b[tot-i+1]=a[i];
	bb=0;
	for (i=1;i<=tot;i++) if (b[i]<a[i]) {
		bb=1; break;
	}
	else if (b[i]>a[i]) break; 
	s=0;
	if (!bb) {
		for (i=1;i<=tot;i++) {
			Ans[s+a[i]+1]=s+1;
			for (j=1;j<=a[i];j++) Ans[s+j]=s+j+1;
			s=s+a[i]+1;
		}
	}
	else {
		for (i=1;i<=tot;i++) {
			Ans[s+b[i]+1]=s+1;
			for (j=1;j<=b[i];j++) Ans[s+j]=s+j+1;
			s=s+b[i]+1;
		}
	}
	for (i=1;i<=n;i++) printf("%d ",Ans[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<bits/stdc++.h>
#define N 100005
#define M 200005
using namespace std;
int i,j,k,l,s,n,m,last[N],to[M],Next[M],a[N],b[N],bb,fa[N],vis[N],Ans[N],ans,ans1,x,y;
int q[N],tot,X,Y;
inline void add(int x,int y) {
	Next[++tot]=last[x]; last[x]=tot; to[tot]=y;
}
inline void dfs(int x) {
	vis[x]=1;
	s++;
	if (s>ans) ans=s,ans1=x;
	for (int i=last[x];i;i=Next[i]) if (!vis[to[i]]) dfs(to[i]);
	vis[x]=0,s--;
}
inline void build(int x) {
	int l=0,r=1; q[1]=x;
	while (l<r) {
		int k=q[++l];
		for (int i=last[k];i;i=Next[i]) if (fa[k]!=to[i]) fa[q[++r]=to[i]]=k;
	}
}
int main() {
	scanf("%d",&n);
	for (i=1;i<n;i++) scanf("%d%d",&x,&y),add(x,y),add(y,x);
	dfs(1);
	ans=0;
	X=ans1;
	dfs(ans1);
	Y=ans1;
	build(X);
	for (i=Y;i;i=fa[i]) vis[i]=1;
	for (i=1;i<=n;i++) if (!vis[i]&&!vis[fa[i]]) return puts("-1"),0;
	tot=0;
	for (i=Y;i;i=fa[i]) {
		tot++;
		for (j=last[i];j;j=Next[j]) {
			if (to[j]==fa[i]) continue;
			if (!vis[to[j]]) a[tot]++;
		}
	}
	for (i=1;i<=tot;i++) b[tot-i+1]=a[i];
	bb=0;
	for (i=1;i<=tot;i++) if (b[i]<a[i]) {
		bb=1; break;
	}
	else if (b[i]>a[i]) break; 
	s=0;
	if (!bb) {
		for (i=1;i<=tot;i++) {
			Ans[s+a[i]+1]=s+1;
			for (j=1;j<=a[i];j++) Ans[s+j]=s+j+1;
			s=s+a[i]+1;
		}
	}
	else {
		for (i=1;i<=tot;i++) {
			Ans[s+b[i]+1]=s+1;
			for (j=1;j<=b[i];j++) Ans[s+j]=s+j+1;
			s=s+b[i]+1;
		}
	}
	for (i=1;i<=n;i++) printf("%d ",Ans[i]);
}
```