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.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Scanner;

public class Main {
	
	public static void main(String args[]) {
		Scanner scan=new Scanner(System.in);
		int n=scan.nextInt();
		int m=scan.nextInt();
		ArrayList<Integer>[] adj=new ArrayList[n];
		for(int i=0;i<n;i++) {
			adj[i]=new ArrayList<Integer>();  
		}
		for(int i=0;i<m;i++) {
			int u=scan.nextInt();
			int v=scan.nextInt();
			u--;v--;
			adj[u].add(v);
			adj[v].add(u);
			
		}
		for(int i=0;i<n;i++) {
			Collections.sort(adj[i]);
		}
		HashMap<String,ArrayList<Integer>> map=new HashMap<String,ArrayList<Integer>>();
		
		
		for(int i=0;i<n;i++) {
			int p[]=new int[adj[i].size()];
			if(adj[i].size()==0) {
				System.out.println(-1);
				return;
			}
			for(int j=0;j<adj[i].size();j++) {
				p[j]=adj[i].get(j);
			}
			String ep=Arrays.toString(p);
//			System.out.println(ep);
			if(map.containsKey(ep)) {
				map.get(ep).add(i);
			}else {
				ArrayList<Integer> x=new ArrayList<Integer>();
				x.add(i);
				map.put(ep,x);
			}
		}
//		System.out.println(map.size());
		if(map.size()==3) {
//			ArrayList<Integer> qq=map.get
			int count=1;
			int ans[]=new int[n+1];
			for(String p:map.keySet()) {
//				count=1;
				for(int a:map.get(p)) {
					
//					System.out.print((a+1)+" ");
					ans[a]=count;
				}
				count++;
				
			}
			for(int i=0;i<n;i++) {
				System.out.print(ans[i]+" ");
			}
		}else {
			System.out.println(-1);
		}
	}
	
}
