Problem: One day Vasya came up to the blackboard and wrote out n distinct integers from 1 to n in some order in a circle. Then he drew arcs to join the pairs of integers (a, b) (a ≠ b), that are either each other's immediate neighbors in the circle, or there is number c, such that a and с are immediate neighbors, and b and c are immediate neighbors. As you can easily deduce, in the end Vasya drew 2·n arcs.

For example, if the numbers are written in the circle in the order 1, 2, 3, 4, 5 (in the clockwise direction), then the arcs will join pairs of integers (1, 2), (2, 3), (3, 4), (4, 5), (5, 1), (1, 3), (2, 4), (3, 5), (4, 1) and (5, 2).

Much time has passed ever since, the numbers we wiped off the blackboard long ago, but recently Vasya has found a piece of paper with 2·n written pairs of integers that were joined with the arcs on the board. Vasya asks you to find the order of numbers in the circle by these pairs.

Input

The first line of the input contains a single integer n (5 ≤ n ≤ 105) that shows, how many numbers were written on the board. Next 2·n lines contain pairs of integers ai, bi (1 ≤ ai, bi ≤ n, ai ≠ bi) — the numbers that were connected by the arcs.

It is guaranteed that no pair of integers, connected by a arc, occurs in the input more than once. The pairs of numbers and the numbers in the pairs are given in the arbitrary order.

Output

If Vasya made a mistake somewhere and there isn't any way to place numbers from 1 to n on the circle according to the statement, then print a single number "-1" (without the quotes). Otherwise, print any suitable sequence of n distinct integers from 1 to n. 

If there are multiple solutions, you are allowed to print any of them. Specifically, it doesn't matter which number you write first to describe the sequence of the order. It also doesn't matter whether you write out the numbers in the clockwise or counter-clockwise direction.

Examples

Input

5
1 2
2 3
3 4
4 5
5 1
1 3
2 4
3 5
4 1
5 2


Output

1 2 3 4 5 

Input

6
5 6
4 3
5 3
2 4
6 1
3 1
6 2
2 5
1 4
3 6
1 2
4 5


Output

1 2 4 5 3 6 

Solution (in JAVA):

Solution:
import java.util.*;
import java.math.*;
import java.io.*;

import static java.lang.Math.*;
import static java.util.Arrays.*;
import static java.util.Collections.*;

public class  test{
	
	// ArrayList<Integer> lis = new ArrayList<Integer>();
	// ArrayList<String> lis = new ArrayList<String>();
	// PriorityQueue<Integer> que = new PriorityQueue<Integer>();
	//  Stack<Integer> que = new Stack<Integer>();
    //	static long sum=0;
	// 1000000007 (10^9+7)
	static int mod = 1000000007;
	//static int mod = 1000000009,r=0;
   // static int dx[]={1,-1,0,0};
//	static int dy[]={0,0,1,-1};
//	static int dx[]={1,-1,0,0,1,1,-1,-1};
//  static int dy[]={0,0,1,-1,1,-1,1,-1};
	//static long H,L;
	//static Set<Integer> set = new HashSet<Integer>();
public  static void main(String[] args)   throws Exception, IOException{
   //String line=""; throws Exception, IOException
   //(line=br.readLine())!=null
	//Scanner sc =new Scanner(System.in);
	// !!caution!! int long //  
	Reader sc = new Reader(System.in);
  // while( ){
	  // int  n=sc.nextInt(),m=sc.nextInt();//a[]=new int[n],b[]=new int[n];
	int  n=sc.nextInt(),x[][]=new int[n+1][5],f=0;
	for(int i=1;i<=n;i++)x[i][0]=1;
	 for(int i=0;i<2*n;i++){
		 int a=sc.nextInt(),b=sc.nextInt();
		 if(0<f || 4<x[a][0] || 4<x[b][0] ){f++;continue;}
		x [a][x[a][0]]=b;
		x [b][x[b][0]]=a;
		x[a][0]++;
		x[b][0]++;
		
		
	 }
	 if(0<f){System.out.println(-1);return;}
//	db(x);
	int y[]=new int[5],z[]=new int[5];
	for(int i=1;i<5;i++)z[i]=x[1][i];z[0]=1;
	
lp:	for(int[] p : new Permutation(5)) {
		//db(p);
		for(int i=0;i<5;i++)y[i]=z[p[i]];
		if( con( x[y[2]],y[1] ) && con( x[y[2]],y[3] ) && con( x[y[0]],y[2] ) && con( x[y[2]],y[4] ) && con( x[y[1]],y[3] ) ){

			boolean b[]=new boolean[n+1];
			int an[]=new int[n];
			for(int i=0;i<5;i++){an[i]=y[i]; b[y[i]]=true; }
			for(int i=5;i<n;i++){
				
				for(int t=1;t<=4;t++){
					if( x[y[3]][t]!=y[1] &&x[y[3]][t]!=y[2] &&x[y[3]][t]!=y[4]  ){
						if( b[x[y[3]][t]] )continue lp;
						b[x[y[3]][t] ]=true;
						an[i]=x[y[3]][t]; int e=x[y[3]][t];
					  y[1]=y[2]; y[2]=y[3];y[3]=y[4]; y[4]=e;	
						break;
					}
				}
			}
			if( !con( x[an[n-1]],an[0] ) || !con( x[an[n-1]],an[1] ) )continue;
			for(int i=0;i<n;i++)System.out.print(an[i]+" ");
		  return;	
		}
		
	}
     
   // db(s,a);
    System.out.println(-1);
    
     
}
     
 static boolean con(int x[],int y){
	 
	 for(int i=1;i<=4;i++)if( x[i]==y )return true;
	 return false ;
 }

static void db(Object... os){
    System.err.println(Arrays.deepToString(os));

}
}



class Reader
{
	private BufferedReader x;
	private StringTokenizer st;
	
	public Reader(InputStream in)
	{
		x = new BufferedReader(new InputStreamReader(in));
		st = null;
	}
	public String nextString() throws IOException
	{
		while( st==null || !st.hasMoreTokens() )
			st = new StringTokenizer(x.readLine());
		return st.nextToken();
	}
	public int nextInt() throws IOException
	{
		return Integer.parseInt(nextString());
	}
	public long nextLong() throws IOException
	{
		return Long.parseLong(nextString());
	}
	public double nextDouble() throws IOException
	{
		return Double.parseDouble(nextString());
	}
}


class Permutation implements Iterable<int[]> {
	private final int size;

	public Permutation(int size) {
		if (size <= 0) throw new IllegalArgumentException();
		this.size = size;
	}

	public Iterator<int[]> iterator() {
		return new Iter(size);
	}

	private class Iter implements Iterator<int[]> {
		private final int[] source;
		private boolean isFirst = true;

		private Iter(int size) {
			source = new int[size];
			for(int i = 0; i < size; ++i) {
				source[i] = i;
			}
		}
		
		public boolean hasNext() {
			if (isFirst) {
				isFirst = false;
				return true;
			}

			int n = source.length;
			for (int i = n - 1; i > 0; i--) {
				if (source[i - 1] < source[i]) {
					int j = n;
					while (source[i - 1] >= source[--j]);
					swap(source, i - 1, j);
					reverse(source, i, n);
					return true;
				}
			}
			reverse(source, 0, n);
			return false;
		}

		public int[] next() {
			return source;
		}

		public void remove() {
			throw new UnsupportedOperationException();
		}

		private void swap(int[] is, int i, int j) {
			int t = is[i];
			is[i] = is[j];
			is[j] = t;
		}

		private void reverse(int[] is, int s, int t) {
			while (s < --t) swap(is, s++, t);
		}
	}
}
/*

class P implements Comparable<P>{
//	implements Comparable<P>
	int v,w;
	P(int v,int w){
		this.v=v;
		this.w=w;
	
	} 
	public int compareTo(P x){
	   	 return w-x.w;//ascend 
	   }
}
*/