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.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashSet;
import java.util.List;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Scanner;
import java.util.Set;
import java.util.SortedMap;
import java.util.SortedSet;
import java.util.TreeMap;
import java.util.TreeSet;

public class Main {

	static SortedMap<Integer, Integer> map = new TreeMap<Integer, Integer>();
	static SortedSet<Integer>[] adj;

	public static void main(String args[]) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt(), m = sc.nextInt();
		adj = new TreeSet[n];
		for (int i = 0; i < n; i++) {
			adj[i] = new TreeSet<Integer>();
		}
		for (int i = 0; i < m; i++) {
			int a = sc.nextInt();
			int b = sc.nextInt();
			adj[a - 1].add(b - 1);
			adj[b - 1].add(a - 1);
		}
		if(m<3) {
			System.out.println(-1);
		}else {
		Set<Integer> size = new HashSet<Integer>();
		for (int i = 0; i < n; i++) {
			size.add(adj[i].size());
		}
		if (size.size() >= 4) {
			System.out.println(-1);
		} else {
			int flag=0;
			TreeSet<Integer> one = new TreeSet();  ArrayList<Integer>three = new ArrayList();
			TreeSet<Integer>two = new TreeSet();
			
				
			int inserted[] = new int[n];
			for(int i=0;i<n;i++) {
				inserted[i]=1;
				one.add(i);
			}
			
			for (int p : adj[0]) {
				    one.remove(p);
					two.add(p);
					inserted[p] = 2;
					
				
			}
			int j=1;
			while(j<n&&inserted[j]==1) {
				j++;
			}
			int tt=-1;
			if(j<n) {
				for (int p : adj[j]) {
					if(inserted[p]!=1) {
						tt=tt==-1?p:tt;
						three.add(p);
						inserted[p] = 3;
						two.remove(p);
					}
					
				}
			}else {
				System.out.println(-1);
			}
			ArrayList<Integer> one1=new ArrayList<Integer>();
			one1.addAll(one);
			
			int x=one.size();

			
				ArrayList<Integer> cc= new ArrayList();
				cc.addAll(adj[0]);
				int ss=adj[0].size();
					for(int bb=0;bb<x;bb++) {
						int g=0;
						if(ss!=adj[one1.get(bb)].size()) {
							flag=2;
							continue;
						}
						for(int hh:adj[one1.get(bb)]) {
						if(hh!=cc.get(g)) {
							flag=3;
						}
						g++;
						}
					}

			ArrayList<Integer> two1=new ArrayList<Integer>();
			two1.addAll(two);
				int x2=two1.size();

				ArrayList<Integer> cc1= new ArrayList();
				cc1.addAll(adj[j]);
					ss=adj[j].size();
						for(int bb=0;bb<x2;bb++) {
							int g=0;
							if(ss!=adj[two1.get(bb)].size()) {
								flag=4;
								break;
							}
							for(int hh:adj[two1.get(bb)]) {
							if(hh!=cc1.get(g)) {
								flag=5;
							}
							g++;
							}
					}
				
					int x3=three.size();

					if(tt!=-1) {

						ArrayList<Integer> cc2= new ArrayList();
						cc2.addAll(adj[tt]);
						ss=adj[tt].size();
							for(int bb=0;bb<x3;bb++) {
								int g=0;
								if(ss!=adj[three.get(bb)].size()) {
									flag=6;
									break;
								}
								for(int hh:adj[three.get(bb)]) {
								if(hh!=cc2.get(g)) {
									flag=7;
								}
								g++;
								}
						}
					}
			if(flag!=0||x==0||x2==0||x3==0||(x+x2+x3)!=n) {
				if(n==100000) {
					
					System.out.println(flag+ "  " +x+ " "+x2+ " "+x3);
				}
				System.out.println(-1);
				
			}else {
				int out[]=new int[n];
				for(int y:one) {
					out[y]=1;
				}
				for(int y:two) {
					out[y]=2;
				}
				for(int y:three) {
					out[y]=3;
				}
				for(int i=0;i<n;i++) {
					System.out.print(out[i]+ " ");
				}
			}
			
		}
	}}

}
