//#include <bits/stdc++.h>
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <limits>
#include <map>
#include <vector>
#include <queue> 
#define LL long long
#define ft first
#define sd second
#define mp(x,y) make_pair(x,y)
//#define int LL
#define db double
using namespace std;
const int N   = 1e5+233;
//const int mod = ;
const int INF =numeric_limits<int >::max();
#define For(n) for (int i=1;i<=n;++i)
#define rep(i,x,y) for (int i=x;i<=y;++i)
void read(int &x)
{
	x=0;
	char ch=getchar();
	int f=1;
	while (!isdigit(ch)) (ch=='-'?f=-1:0),ch=getchar();
	while ( isdigit(ch)) x=x*10+ch-'0',ch=getchar();
	x*=f;
}
struct xx
{
	int V,next;
}b[N<<1];
int G,head[N],vis[N],dep[N],mx;
void add(int x,int y)
{
	b[++G]=(xx){y,head[x]};
	head[x]=G;
}
int n,x,y,rt,fa[N];
#define F(x) for (int i=head[x];i;i=b[i].next)
#define v b[i].V
void dfs1(int u,int Fa=0)
{
	if (dep[u]>dep[mx]) mx=u;
	F(u) if (v^Fa) fa[v]=u,dep[v]=dep[u]+1,dfs1(v,u);
}
int son[N],st[N],tp;
void dfs2(int u,int Fa=0)
{
	F(u) if (v^Fa) 
	{
		if (!vis[v]) dep[v]=dep[u]+1,son[u]++;
		dfs2(v,u);
	}
}
int k[N],ans1[N],ans2[N],ok=1;

void solve(int * p)
{
	int nw=0;
	rep(i,1,tp) 
	{
		p[nw+k[i]+1]=nw+1;
		rep(j,1,k[i]) p[nw+j]=nw+j+1;
		nw+=k[i]+1;
	}
}

signed main()
{
	read(n);
	rep(i,2,n) read(x),read(y),add(x,y),add(y,x);
	mx=1;dfs1(1);//
	dep[rt=mx]=0;dfs1(mx);
	for (int k=mx;k^rt;k=fa[k]) vis[st[++tp]=k]=1;vis[st[++tp]=rt]=1;
	memset(dep,0,sizeof(dep));
	dfs2(rt);
	rep(i,1,n) if (dep[i]>1) return printf("-1"),0;
	rep(i,1,tp) k[i]=son[st[i]];
	solve(ans1);
	rep(i,1,tp/2) swap(k[i],k[tp+1-i]);
	solve(ans2);
	rep(i,1,n) if (ans1[i]!=ans2[i])
	{
		if (ans1[i]<ans2[i]) ok=1;
		else ok=2;
		break;
	}
	if (ok==1) rep(i,1,n) printf("%d ",ans1[i]);
	else rep(i,1,n) printf("%d ",ans2[i]);
	return 0; 
}

### Prompt

Develop a solution in Cpp 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>
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <limits>
#include <map>
#include <vector>
#include <queue> 
#define LL long long
#define ft first
#define sd second
#define mp(x,y) make_pair(x,y)
//#define int LL
#define db double
using namespace std;
const int N   = 1e5+233;
//const int mod = ;
const int INF =numeric_limits<int >::max();
#define For(n) for (int i=1;i<=n;++i)
#define rep(i,x,y) for (int i=x;i<=y;++i)
void read(int &x)
{
	x=0;
	char ch=getchar();
	int f=1;
	while (!isdigit(ch)) (ch=='-'?f=-1:0),ch=getchar();
	while ( isdigit(ch)) x=x*10+ch-'0',ch=getchar();
	x*=f;
}
struct xx
{
	int V,next;
}b[N<<1];
int G,head[N],vis[N],dep[N],mx;
void add(int x,int y)
{
	b[++G]=(xx){y,head[x]};
	head[x]=G;
}
int n,x,y,rt,fa[N];
#define F(x) for (int i=head[x];i;i=b[i].next)
#define v b[i].V
void dfs1(int u,int Fa=0)
{
	if (dep[u]>dep[mx]) mx=u;
	F(u) if (v^Fa) fa[v]=u,dep[v]=dep[u]+1,dfs1(v,u);
}
int son[N],st[N],tp;
void dfs2(int u,int Fa=0)
{
	F(u) if (v^Fa) 
	{
		if (!vis[v]) dep[v]=dep[u]+1,son[u]++;
		dfs2(v,u);
	}
}
int k[N],ans1[N],ans2[N],ok=1;

void solve(int * p)
{
	int nw=0;
	rep(i,1,tp) 
	{
		p[nw+k[i]+1]=nw+1;
		rep(j,1,k[i]) p[nw+j]=nw+j+1;
		nw+=k[i]+1;
	}
}

signed main()
{
	read(n);
	rep(i,2,n) read(x),read(y),add(x,y),add(y,x);
	mx=1;dfs1(1);//
	dep[rt=mx]=0;dfs1(mx);
	for (int k=mx;k^rt;k=fa[k]) vis[st[++tp]=k]=1;vis[st[++tp]=rt]=1;
	memset(dep,0,sizeof(dep));
	dfs2(rt);
	rep(i,1,n) if (dep[i]>1) return printf("-1"),0;
	rep(i,1,tp) k[i]=son[st[i]];
	solve(ans1);
	rep(i,1,tp/2) swap(k[i],k[tp+1-i]);
	solve(ans2);
	rep(i,1,n) if (ans1[i]!=ans2[i])
	{
		if (ans1[i]<ans2[i]) ok=1;
		else ok=2;
		break;
	}
	if (ok==1) rep(i,1,n) printf("%d ",ans1[i]);
	else rep(i,1,n) printf("%d ",ans2[i]);
	return 0; 
}

```