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.*;
import java.io.*;

public class Solution1{

	public static Integer INT(String s){
		return Integer.parseInt(s);
	}

	public static Long LONG(String s){
		return Long.parseLong(s);
	}

	//====================================================================================================================


	static HashSet<Integer> adj[];
	static int n, m, type[];


	public static void main(String args[])throws IOException{

		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));	Scanner in=new Scanner(System.in); StringBuilder out=new StringBuilder();

		n=in.nextInt();
		m=in.nextInt();

		adj=new HashSet[n];
		for(int i=0; i<n; i++)
			adj[i]=new HashSet<>();

		for(int i=0; i<m; i++){
			int u=in.nextInt()-1,
				v=in.nextInt()-1;

			adj[u].add(v); 
			adj[v].add(u);
		}

		type=new int[n];
		Arrays.fill(type, -1);


		if(!solve()) {
			System.out.println(-1);
			return;
		}

		for(int i=0; i<n; i++)
			out.append((type[i]+1)+" ");
		System.out.println(out);
	}

	public static boolean solve() {

		dfs(0, -1, -1);

		for(int item: type)
			if(item==-1)
				return false;

		for(int i=0; i<n; i++)
			for(int item: adj[i])
				if(type[i]==type[item])
					return false;

		int freq[]=new int[3], sum=0;
		for(int item: type)
			freq[item]+=1;

		for(int i=0; i<=2; i++)
			if(freq[i]==0)
				return false;
			else
				sum+=freq[i];

		for(int i=0; i<n; i++) {
			if(adj[i].size()!=(sum-freq[type[i]]))
				return false;
		}
		
		return true;
	}

	public static void dfs(int i, int p, int gp) {
		
		if(gp!=-1 && !adj[i].contains(gp))
			type[i]=type[gp];
		else {
			boolean flag[]=new boolean[3];
			if(p!=-1)
				flag[type[p]]=true;
			if(gp!=-1)
				flag[type[gp]]=true;
			for(int k=0; k<=2; k++)
				if(!flag[k]) {
					type[i]=k;
					break;
				}
		}

		for(int item: adj[i]) 
			if(type[item]==-1)
				dfs(item, i, p);
	}
}