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.InputStreamReader;
import java.io.PrintWriter;
import java.lang.reflect.Array;
import java.util.*;

public class F {
    public static class Gragh {
        public ArrayList<Integer>[] edges;
        public Gragh(int size) {
            this.edges = new ArrayList[size];
            for (int i = 0; i < size; ++i) {
                edges[i] = new ArrayList<>();
            }
        }

        public void add(int x, int y) {
            edges[x].add(y);
        }

        public ArrayList<Integer> getn(int x) {
            return edges[x];
        }
    }
    public static Gragh gragh;
    public static int[] part;
    public static boolean[] used;
    public static boolean ok = true;
    public static void dfs(int node){
        used[node] = true;
        ArrayList<Integer> nn = gragh.getn(node);
        for(Integer n : nn){
            if(part[n] == part[node]){
                ok = false;
            }
            if(!used[n]){
                dfs(n);
            }
        }
    }
    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter writer = new PrintWriter(System.out);
        StringTokenizer st = new StringTokenizer(reader.readLine());
        int n = Integer.parseInt(st.nextToken());
        int m = Integer.parseInt(st.nextToken());
        gragh = new Gragh(n);
        for(int i = 0; i < m; ++i){
            st = new StringTokenizer(reader.readLine());
            int x = Integer.parseInt(st.nextToken()) - 1;
            int y = Integer.parseInt(st.nextToken()) - 1;
            gragh.add(x, y);
            gragh.add(y, x);
        }
        part = new int[n];
        used = new boolean[n];
        for(int i = 0; i < n; ++i){
            used[i] = true;
        }
        ArrayList<Integer> nn = gragh.getn(0);
        for(Integer node : nn){
            used[node] = false;
        }
        int[] c = new int[3];
        for(int i = 0; i < n; ++i){
            if(!used[i]){
                nn = gragh.getn(i);
                for(Integer node : nn){
                    if(!used[node]) {
                        used[node] = true;
                        part[node] = 3;
                        ++c[2];
                    }
                }
                break;
            }
        }

        for(int i = 0; i < n; ++i){
            if(!used[i]){
                part[i] = 2;
                ++c[1];
            }
            else if(used[i] && part[i] == 0){
                part[i] = 1;
                ++c[0];
            }
        }
        if(c[0] == 0 || c[1] == 0 || c[2] == 0){
            System.out.println(-1);
            return;
        }
        used = new boolean[n];
        dfs(0);
        if(!ok){
            System.out.println(-1);
            return;
        }
        for(int i = 0; i < n; ++i){
            if(gragh.getn(i).size() != n - c[part[i] - 1]){
                System.out.println(-1);
                return;
            }
            if(!used[i]){
                System.out.println(-1);
                return;
            }
        }
        for(int i = 0; i < n; ++i){
            writer.print(part[i] + " ");
        }
        writer.close();
    }
}