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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;

/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */
/**
 *
 * @author Andy Phan
 */
public class d {
    static ArrayList<Integer>[] edges;
    static int n;
    static int[] set;

    public static void main(String[] args) {
        FS in = new FS(System.in);
        PrintWriter out = new PrintWriter(System.out);
        
        int n = in.nextInt();
        int m = in.nextInt();
        ArrayList<Integer>[] edges = new ArrayList[n];
        Arrays.setAll(edges, (i) -> new ArrayList<Integer>());
        for(int i = 0; i < m; i++) {
            int u = in.nextInt()-1;
            int v = in.nextInt()-1;
            edges[u].add(v);
            edges[v].add(u);
        }
        
        int[] set = new int[n];
        int[] count = new int[3];
        int tmp = -1;
        for(int i : edges[0]) {
            set[i] = 1;
            tmp = i;
            count[1]++;
        }
        
        if(tmp == -1) {
            System.out.println(-1);
            return;
        }
        int tmp2 = -1;
        for(int i : edges[tmp]) {
            if(set[i] == 0) continue;
            set[i] = 2;
            tmp2 = i;
            count[1]--;
            count[2]++;
        }
        count[0] = n-count[1]-count[2];
        if(tmp2 == -1) {
            System.out.println(-1);
            return;
        }
        for(int i = 0; i < n; i++) for(int j : edges[i]) if(set[i] == set[j]) {System.out.println(-1); return;}
        
        if(num(n)-m != num(count[0])+num(count[1])+num(count[2])) {System.out.println(-1); return;}
        for(int i : set) out.print((i+1) + " ");
        out.println("");
        out.close();
    }
    
    static long num(int n) {
        return (long)n*(n-1)/2;
    }

    static class FS {

        BufferedReader in;
        StringTokenizer token;

        public FS(InputStream str) {
            in = new BufferedReader(new InputStreamReader(str));
        }

        public String next() {
            if (token == null || !token.hasMoreElements()) {
                try {
                    token = new StringTokenizer(in.readLine());
                } catch (IOException ex) {
                }
                return next();
            }
            return token.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }
    }
}
