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.
// package Quarantine;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashSet;
import java.util.StringTokenizer;

public class Tripartite {
    public static void main(String[] args)throws IOException {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer(br.readLine());
        int n=Integer.parseInt(st.nextToken());
        int m=Integer.parseInt(st.nextToken());
        ArrayList<Integer> graph[]=new ArrayList[n+1];
        for(int i=1;i<=n;i++){
            graph[i]=new ArrayList<>();
        }
        for(int i=1;i<=m;i++){
            st=new StringTokenizer(br.readLine());
            int x=Integer.parseInt(st.nextToken());
            int y=Integer.parseInt(st.nextToken());
            graph[x].add(y);
            graph[y].add(x);
        }
        HashSet<Integer> set=new HashSet<>();
        for(int i=1;i<=n;i++){
            Collections.sort(graph[i]);
            set.add(graph[i].hashCode());
        }
        StringBuilder print=new StringBuilder();
        if(set.size()==3){
            HashSet<Integer> sets[]=new HashSet[3];
            for(int i=0;i<3;i++){
                sets[i]=new HashSet<>();
            }
            int ind=0;
            for(int k:set){
                for(int i=1;i<=n;i++){
                    if(graph[i].hashCode()==k){
                        sets[ind].add(i);
                    }
                }
                ind++;
            }
            for(int i=1;i<=n;i++){
                if(sets[0].contains(i)){
                    if(graph[i].size()!=sets[1].size()+sets[2].size()){
                        System.out.println(-1);
                        return;
                    }
                    print.append("1 ");
                }
                else if(sets[1].contains(i)){
                    if(graph[i].size()!=sets[2].size()+sets[0].size()){
                        System.out.println(-1);
                        return;
                    }
                    print.append("2 ");
                }
                else{
                    if(graph[i].size()!=sets[1].size()+sets[0].size()){
                        System.out.println(-1);
                        return;
                    }
                    print.append("3 ");
                }
            }
            System.out.println(print.toString());
        }
        else{
            System.out.println(-1);
        }
    }
}
