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.
/*
If you want to aim high, aim high
Don't let that studying and grades consume you
Just live life young
*/
import java.util.*;
import java.io.*;

   public class D
   {
      public static void main(String args[]) throws Exception
      {
         BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));  
         StringTokenizer st = new StringTokenizer(infile.readLine());
         int N = Integer.parseInt(st.nextToken());
         int M = Integer.parseInt(st.nextToken());
         LinkedList<Integer>[] edges = new LinkedList[N+1];
         for(int i=1; i <= N; i++)
            edges[i] = new LinkedList<Integer>();
         for(int i=0; i < M; i++)
         {
            st = new StringTokenizer(infile.readLine());
            int a = Integer.parseInt(st.nextToken());
            int b = Integer.parseInt(st.nextToken());
            edges[a].add(b);
            edges[b].add(a);
         }
         //if two nodes are not adjacent, they must belong in the same set
         //so let's create these three groups
         HashSet<Integer>[] groups = new HashSet[3];
         int[] res = new int[N+1];
         for(int i=0; i < 3; i++)
            groups[i] = new HashSet<Integer>();
         groups[0].add(1);
         boolean[] seen = new boolean[N+1];
         for(int next: edges[1])
            seen[next] = true;
         for(int i=2; i <= N; i++)
            if(!seen[i])
               groups[0].add(i);
         //first group done
         //let's find a 2nd group
         for(int i=2; i <= N; i++)
            if(seen[i]) //if not in first group
            {
               //create 2nd and 3rd groups
               groups[1].add(i);
               boolean[] seen2 = new boolean[N+1];
               seen2[i] = true;
               for(int next: edges[i])
               {
                  if(!seen[next])
                     continue;
                  groups[2].add(next);
                  seen2[next] = true;
               }
               for(int j=2; j <= N; j++)
                  if(seen[j] && !seen2[j])
                     groups[1].add(j);
               break;
            }
         //last step is to check if the tripartite conditions hold
         //each edge has to be one to one
         //this *might* be slow
         for(int i=0; i < 3; i++)
            for(int first: groups[i])
            {
               if(edges[first].size() != N-groups[i].size())
               {
                  System.out.println(-1);
                  return;
               }
               res[first] = i;
            }
         //bfs
         Queue<Integer> q = new LinkedList<Integer>();
         q.add(1);
         seen = new boolean[N+1];
         seen[1] = true;
         while(q.size() > 0)
         {
            int curr = q.poll();
            for(int next: edges[curr])
            {
               if(res[curr] == res[next])
               {
                  System.out.println(-1);
                  return;
               }
               if(!seen[next])
               {
                  seen[next] = true;
                  q.add(next);
               }
            }
         }
         boolean works = true;
         HashSet<Integer> boof = new HashSet<Integer>();
         for(int i=0; i < 3; i++)
            boof.add(i);
         for(int i=1; i <= N; i++)
         {
            works &= seen[i];
            if(boof.contains(res[i]))
               boof.remove(res[i]);
         }
         works &= boof.size() == 0;
         if(works)
         {
            StringBuilder sb = new StringBuilder();
            for(int i=1; i <= N; i++)
               sb.append((1+res[i])+" ");
            System.out.println(sb.toString());
         }
         else
            System.out.println(-1);
      }
   }