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
#include<bits/stdc++.h>
using namespace std;

bool bo[100010];
int n,hd[100010],cnt,x,y,d[100010],p[100010],po[100010],ans[100010],q[100010],l,r;
struct node{int to,next;}e[200010];

void addedge(int x,int y)
{
	e[++cnt]=(node){y,hd[x]},hd[x]=cnt;
	e[++cnt]=(node){x,hd[y]},hd[y]=cnt;
	d[x]++,d[y]++;
}

int main()
{
	scanf("%d",&n);
	if (n==2) return puts("1 2"),0;
	for (int i=1; i<n; i++) scanf("%d%d",&x,&y),addedge(x,y);
	for (int i=1; i<=n; i++) if (d[i]>1) cnt++,q[l=r=1]=i;
	while (l<=r)
	{
		x=q[l++],bo[x]=1;
		for (int i=hd[x]; i; i=e[i].next)
			if (!bo[e[i].to]&&d[e[i].to]>1) q[++r]=e[i].to;
	}
	if (r!=(cnt-=(2*n-2))) return puts("-1"),0;
	q[1]=q[r],l=r=1,memset(bo,0,sizeof(bo));
	while (l<=r)
	{
		x=q[l++],bo[x]=1,cnt=0;
		for (int i=hd[x]; i; i=e[i].next)
			if (!bo[e[i].to]&&d[e[i].to]>1) cnt++,q[++r]=e[i].to;
		if (cnt>1) return puts("-1"),0;
	}
	p[cnt=1]=1;
	for (int i=1,nw=1; i<=r; i++)
	{
		p[cnt+1]=nw+d[q[i]]-1,cnt++;
		for (int j=nw+1; j<nw+d[q[i]]-1; j++) p[++cnt]=j;
		nw+=d[q[i]]-1;
	}
	p[++cnt]=n;
	for (int i=1; i<=n; i++) ans[p[i]]=i;
	reverse(q+1,q+1+r);
	p[cnt=1]=1;
	for (int i=1,nw=1; i<=r; i++)
	{
		p[cnt+1]=nw+d[q[i]]-1,cnt++;
		for (int j=nw+1; j<nw+d[q[i]]-1; j++) p[++cnt]=j;
		nw+=d[q[i]]-1;
	}
	p[++cnt]=n;
	for (int i=1; i<=n; i++) po[p[i]]=i;
	for (int i=1; i<=n; i++)
		if (po[i]!=ans[i])
		{
			if (po[i]<ans[i]) memcpy(ans,po,sizeof(ans));
			break;
		}
	for (int i=1; i<=n; i++) printf("%d ",ans[i]);
	return puts(""),0;
}