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.*;

public class d589 {
	public static void main(String[] args) {
		Scanner s=new Scanner(System.in);
		int num_nodes=s.nextInt();
		int num_edges=s.nextInt();
		int[][] edges=new int[num_edges][2];
		HashSet<Integer>[] graph=new HashSet[num_nodes];
		for(int i=0;i<num_nodes;i++)graph[i]=new HashSet<Integer>();
		
		for(int i=0;i<num_edges;i++) {
			int start=s.nextInt()-1;
			int end=s.nextInt()-1;
			graph[start].add(end);
			graph[end].add(start);
			int[] edge= {start,end};
			edges[i]=edge;
		}
		s.close();
		HashSet<Integer> set1=new HashSet<Integer>();
		HashSet<Integer> set2=new HashSet<Integer>();
		HashSet<Integer> set3=new HashSet<Integer>();
		
		HashSet<Integer> notSet1=new HashSet<Integer>();
		set1.add(0);
		for(int i=1;i<num_nodes;i++) {
			if(graph[0].contains(i)){
				notSet1.add(i);
			}else {
				set1.add(i);
				
			}
		}
		int set2base=-1;
		for(int x:notSet1) {set2base=x;break;}
		set2.add(set2base);
		notSet1.remove(new Integer(set2base));
		for(int node:notSet1) {
			if(graph[set2base].contains(node)) {
				set3.add(node);
			}else {
				set2.add(node);
				
			}
		}
		if(set2.isEmpty()||set3.isEmpty()) {
			System.out.println(-1);
			return;
		}
		
//		System.out.println(set1);
//		System.out.println(set2);
//		System.out.println(set3);
		long num_edges_expected=(long)set1.size()*(set2.size()+set3.size())+(long)set2.size()*set3.size();
		if(num_edges_expected!=num_edges) {
			System.out.println(-1);
			return;
		}else {
			for(int node:set1) {
				if(graph[node].size()!=set2.size()+set3.size()) {
					System.out.println(-1);
					return;
				}
			}
			for(int node:set2) {
				if(graph[node].size()!=set1.size()+set3.size()) {
					System.out.println(-1);
					return;
				}
			}
			for(int node:set3) {
				if(graph[node].size()!=set1.size()+set2.size()) {
					System.out.println(-1);
					return;
				}
			}
			for(int[] edge:edges) {
				if(set1.contains(edge[0])&&set1.contains(edge[1])) {
					System.out.println(-1);
					return;
				}
				if(set3.contains(edge[0])&&set3.contains(edge[1])) {
					System.out.println(-1);
					return;
				}
				if(set2.contains(edge[0])&&set2.contains(edge[1])) {
					System.out.println(-1);
					return;
				}
			}
		}
		int[] answer=new int[num_nodes];
		for(int x:set1)answer[x]=1;
		for(int x:set3)answer[x]=3;
		for(int x:set2)answer[x]=2;
		
		
		for(int node:answer)System.out.print(node+" ");
		System.out.println();
		
		
		
		
	}
}


