You have a simple undirected graph consisting of n vertices and m edges. The graph doesn't contain self-loops, there is at most one edge between a pair of vertices. The given graph can be disconnected.

Let's make a definition.

Let v_1 and v_2 be two some nonempty subsets of vertices that do not intersect. Let f(v_{1}, v_{2}) be true if and only if all the conditions are satisfied:

  1. There are no edges with both endpoints in vertex set v_1. 
  2. There are no edges with both endpoints in vertex set v_2. 
  3. For every two vertices x and y such that x is in v_1 and y is in v_2, there is an edge between x and y. 



Create three vertex sets (v_{1}, v_{2}, v_{3}) which satisfy the conditions below;

  1. All vertex sets should not be empty. 
  2. Each vertex should be assigned to only one vertex set. 
  3. f(v_{1}, v_{2}), f(v_{2}, v_{3}), f(v_{3}, v_{1}) are all true. 



Is it possible to create such three vertex sets? If it's possible, print matching vertex set for each vertex.

Input

The first line contains two integers n and m (3 ≤ n ≤ 10^{5}, 0 ≤ m ≤ min(3 ⋅ 10^{5}, (n(n-1))/(2))) — the number of vertices and edges in the graph.

The i-th of the next m lines contains two integers a_{i} and b_{i} (1 ≤ a_{i} < b_{i} ≤ n) — it means there is an edge between a_{i} and b_{i}. The graph doesn't contain self-loops, there is at most one edge between a pair of vertices. The given graph can be disconnected.

Output

If the answer exists, print n integers. i-th integer means the vertex set number (from 1 to 3) of i-th vertex. Otherwise, print -1.

If there are multiple answers, print any.

Examples

Input


6 11
1 2
1 3
1 4
1 5
1 6
2 4
2 5
2 6
3 4
3 5
3 6


Output


1 2 2 3 3 3 

Input


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


Output


-1

Note

In the first example, if v_{1} = \{ 1 \}, v_{2} = \{ 2, 3 \}, and v_{3} = \{ 4, 5, 6 \} then vertex sets will satisfy all conditions. But you can assign vertices to vertex sets in a different way; Other answers like "2 3 3 1 1 1" will be accepted as well.

<image>

In the second example, it's impossible to make such vertex sets.
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.Random;
import java.util.StringTokenizer;
import java.util.TreeMap;

public class Solution{
	
	
	
    public static void main(String[] args) throws IOException {
     
    	FastScanner fs = new FastScanner();
    	PrintWriter out = new PrintWriter(System.out);
    	
    		
    		
    	int tt = 1;
    	while(tt-->0) {
    		
    		int n = fs.nextInt(), m = fs.nextInt();
    		HashSet<Integer>[] adjlist = new HashSet[n];
    		for(int i=0;i<n;i++ )adjlist[i] = new HashSet<Integer>();
    		
    		for(int i=0;i<m;i++) {
    			int u = fs.nextInt()-1, v = fs.nextInt()-1;
    			adjlist[u].add(v);
    			adjlist[v].add(u);
    		}
    		
    		//if two vertices are not connected then they must be in same group
    		//making groups
    		
    		int[] g = new int[n];
    		Arrays.fill(g, -1);
    		
    		for(int k=0;k<3;k++) {
    			int first = -1;
    			for(int i=0;i<n;i++) if(g[i]==-1) {first = i; break;}
    			if(first==-1) {
    				out.println(-1);
    				out.flush();
    				return;
    			}
    			
    			g[first] = k;
    			for(int i=0;i<n;i++) {
    				if(g[i]==-1 && !adjlist[first].contains(i)) {
    					g[i] = k;
    				}
    			}
    		}
    		
    		ArrayList<Integer>[] groups = new ArrayList[3];
    		for(int i=0;i<3;i++) groups[i] = new ArrayList<Integer>();
    		
    		for(int i=0;i<n;i++) {
    			if(g[i]==-1) {
    				out.println(-1);
    				out.flush();
    				return;
    			}
    			groups[g[i]].add(i);
    		}
    		
    		
    		//counting the number of edges.
    		int cnt = 0;
    		
    		//it will not be O(n^2) because there is upper limit on the number of edges.
    		for(int g1=0;g1<3;g1++) {
    			for(int g2=g1+1;g2<3;g2++) {
    				for(int v1: groups[g1]) {
    					for(int v2: groups[g2]) {
    						if(adjlist[v1].contains(v2)) cnt++;
    						else {
    							out.println(-1);
    							out.flush();
    							return;
    						}
    					}
    				}
    			}
    		}
    		
    		int x = 1;
    		
    		if(m!=cnt) {
    			out.println(-1);
    			out.flush();
    			return;
    		}
    		
    		for(int i=0;i<n;i++) out.print(g[i]+1+" ");
    		out.println();
    		
    		
    	}
    	
    		
    		
    		
    	out.close();
    		
    }
    
    
   
  
    
    
    
    static final Random random=new Random();
    	
    static void ruffleSort(int[] a) {
    	int n=a.length;//shuffle, then sort 
    	for (int i=0; i<n; i++) {
    		int oi=random.nextInt(n); int temp=a[oi];
    		a[oi]=a[i]; a[i]=temp;
    	}
    	Arrays.sort(a);
    }
   
  
    
    	
    	
    static class FastScanner{
    	BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    	StringTokenizer st = new StringTokenizer("");
     
    	public String next(){
    		while(!st.hasMoreElements()){
    			try{
    				st = new StringTokenizer(br.readLine());
    			} catch(IOException e){
    				e.printStackTrace();
    			}
    		}
    		return st.nextToken();
    	}
    		
    	public String nextLine() throws IOException {
    		return br.readLine();
    	}
    		
    	public int nextInt(){
    		return Integer.parseInt(next());
    	}
     
    	public int[] readArray(int n){
    		int[] a = new int[n];
    		for(int i=0;i<n;i++)
    			a[i] = nextInt();
    		return a;
    	}
    		
    	public long nextLong() {
    		return Long.parseLong(next());
    	}
    		
    	public char nextChar() {
    		return next().toCharArray()[0];
    	}
    }
   	
}
