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.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.util.*;

public class D {

    public Set<Integer> fill(Set<Integer> vertices, Map<Integer, Set<Integer>> g) {
        Set<Integer> vi = new HashSet<>();
        for (int v : vertices) {
            vi.add(v);
            for (int u : vertices) {
                if (!g.get(v).contains(u)) {
                    vi.add(u);
                }
            }
            break;
        }
        for (int v : vi) {
            vertices.remove(v);
        }
        return vi;
    }

    void solve() throws Exception {
        int n = ni();
        int m = ni();
        Map<Integer, Set<Integer>> g = new HashMap<>();
        int[] edge1 = new int[m];
        int[] edge2 = new int[m];
        Set<Integer> vertices = new HashSet<>();

        for (int i = 1; i <= n; i++) {
            g.put(i, new HashSet<>());
            vertices.add(i);
        }
        for (int i = 0; i < m; i++) {
            int u = ni();
            int v = ni();
            g.get(u).add(v);
            g.get(v).add(u);
            edge1[i] = u;
            edge2[i] = v;
        }
        Set<Integer> v1 = fill(vertices, g);
        Set<Integer> v2 = fill(vertices, g);
        Set<Integer> v3 = fill(vertices, g);
        if (v1.size() + v2.size() + v3.size() != n || v1.isEmpty() || v2.isEmpty() || v3.isEmpty()) {
            out.println(-1);
            return;
        }
        for (int i = 0; i < m; i++) {
            int u = edge1[i];
            int v = edge2[i];
            if (v1.contains(u) && v1.contains(v) || v2.contains(u) && v2.contains(v) || v3.contains(u) && v3.contains(v)) {
                out.println(-1);
                return;
            }
        }
        if (v1.size() * v2.size() + v2.size() * v3.size() + v3.size() * v1.size() != m) {
            out.println(-1);
            return;
        }
        for (int i = 1; i <= n; i++) {
            if (v1.contains(i)) {
                out.print("1 ");
            } else if (v2.contains(i)) {
                out.print("2 ");
            } else if (v3.contains(i)) {
                out.print("3 ");
            }
        }
        out.println();
    }

    void run() throws Exception {
        initIO();
        solve();
        out.flush();
    }

    public static void main(String[] args) throws Exception {
        new D().run();
    }

    /**
     * IO.
     */

    PrintWriter out;
    BufferedReader br;
    StringTokenizer tokenizer;

    void initIO() {
        out = new PrintWriter(System.out);
        br = new BufferedReader(new InputStreamReader(System.in));
        tokenizer = null;
    }

    int ni() throws Exception {
        return Integer.parseInt(ns());
    }

    long nl() throws Exception {
        return Long.parseLong(ns());
    }

    double nd() throws Exception {
        return Double.parseDouble(ns());
    }

    int[] nia(int n) throws Exception {
        int[] x = new int[n];
        for (int i = 0; i < x.length; i++) {
            x[i] = ni();
        }
        return x;
    }

    long[] nla(int n) throws Exception {
        long[] x = new long[n];
        for (int i = 0; i < x.length; i++) {
            x[i] = nl();
        }
        return x;
    }

    double[] nda(int n) throws Exception {
        double[] x = new double[n];
        for (int i = 0; i < x.length; i++) {
            x[i] = nd();
        }
        return x;
    }

    String ns() throws Exception {
        while (tokenizer == null || !tokenizer.hasMoreTokens())
            tokenizer = new StringTokenizer(br.readLine());
        return tokenizer.nextToken();
    }

    String nline() throws Exception {
        tokenizer = null;
        return br.readLine();
    }
}
