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 My {
        public static void main(String args[]) {
         Scanner in=new Scanner(System.in);
         int n=in.nextInt();
         int m=in.nextInt();
         HashMap<Integer,ArrayList<Integer>> hm=new HashMap<>();
         for(int i=0;i<m;i++)
         {
             int a=in.nextInt();
             int b=in.nextInt();
             if(!hm.containsKey(a)) hm.put(a,new ArrayList<Integer>());
             if(!hm.containsKey(b)) hm.put(b,new ArrayList<Integer>());
          hm.get(a).add(b);
          hm.get(b).add(a);
         }
        if(hm.size()<n) { System.out.println("-1"); }
        else{
            for(int i=1;i<=n;i++)
            {
                ArrayList<Integer>t=new ArrayList<Integer>();
                t=hm.get(i);
                Collections.sort(t);
                hm.put(i,t);
            }
            
         int w[]=new int[n+1];
         Arrays.fill(w,0);
         boolean visited[]=new boolean[n+1]; Arrays.fill(visited,false);
                 
             
         int cnt=0;
         for(int i=1;i<=n;i++)
         {
             if(cnt==3) break;
             if(!visited[i]){
                 cnt++;
                 w[i]=cnt;
             visited[i]=true;
             ArrayList<Integer> temp1=new ArrayList<Integer>();
             ArrayList<Integer> temp2=new ArrayList<Integer>();
             temp1=hm.get(i);
             
             for(int j=i+1;j<=n;j++)
             { 
                 if(!visited[j]){
                 temp2=hm.get(j);
          
                     if(temp1.equals(temp2)) { visited[j]=true; w[j]=cnt;}
        
             }
             }
             }
         }
         int flag=0;
         if(cnt<3) flag=1;
         for(int i=1;i<=n;i++)
         {
             if(w[i]==0) { flag=1; break; }
         }
         if(flag==0){
         for(int i=1;i<=n;i++)
         {
             System.out.print(w[i]+" ");
         }
         }
         else System.out.println("-1");
        }
           
    }
    }