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.*;
import java.util.*;

public class Main {
    static PrintWriter out = new PrintWriter(System.out);
    static Scanner sc = new Scanner(System.in);
    static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    long mod = 998244353;
    List<Integer>[] graph;
    int size;
    int[] id;
    public static void main(String[] args) throws IOException {
        Main main = new Main();
        main.solve();
//        out.println(main.solve());
        out.flush();
    }
    void solve() throws IOException {
        String[] buf = reader.readLine().split(" ");
        int n = Integer.parseInt(buf[0]), m = Integer.parseInt(buf[1]);
        graph = new List[n+1];
        for(int i=0; i<=n; i++) graph[i] = new LinkedList<>();
        size = n;
        id = new int[n+1];
        for(int i=0;i<=n;i++) id[i] = i;
        for(int i=0;i<m;i++){
            buf = reader.readLine().split(" ");
            int u = Integer.parseInt(buf[0]), v = Integer.parseInt(buf[1]);
            union(u,v);
            graph[u].add(v); graph[v].add(u);
        }
        boolean flag = size==1;
        for(int i=1; i<=n&&flag; i++){
            flag &= graph[i].size()>1;
        }
        if(!flag){
            out.println("-1"); return;
        }
        boolean[] temp = new boolean[n+1],used = new boolean[n+1];
        for(int i:graph[1]) temp[i] = true;
        Set<Integer> set1 = new HashSet<>(n/2+1), set2 = new HashSet<>(n/2+1), set3 = new HashSet<>(n/2+1);
        int s2 = graph[1].get(0);
        for(int i=1;i<=n;i++){
            if(!temp[i]) {
                set1.add(i);
                used[i] = true;
            }
        }
        Arrays.fill(temp,false);
        for(int i:graph[s2]) temp[i] = true;
        set2.add(s2);
        for(int i=1;i<=n;i++){
            if(!temp[i]) {
                set2.add(i);
                used[i] = true;
            }
        }
        for(int i=1;i<=n;i++){
            if(!used[i]) set3.add(i);
            used[i] = true;
        }
        // then check the three sets one by one
        flag = check(set1)&&check(set2)&&check(set3);
        if(!flag){
            out.println("-1");return;
        }
        long n1 = set1.size(), n2 = set2.size(), n3 = set3.size();
        flag = n1*n2+n2*n3+n3*n1==m;
        if(!flag){
            out.println("-1");return;
        }
        int[] res = new int[n+1];
        for(int i:set1) res[i] = 1;
        for(int i:set2) res[i] = 2;
        for(int i:set3) res[i] = 3;
        for(int i=1;i<=n;i++){
            out.print(res[i]);out.print(" ");
        }
    }
    boolean check(Set<Integer> set){
        if(set.size()==0) return false;
        for(int u:set){
            for(int to:graph[u]){
                if(set.contains(to)) return false;
            }
        }
        return true;
    }
    int find(int p){
        while(id[p]!=p){
            id[p] = id[id[p]];
            p = id[p];
        }
        return p;
    }
    void union(int p, int q){
        int i = find(p), j = find(q);
        if(i==j) return;
        id[j] = i;
        size --;
    }
}