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 javafx.beans.binding.When;

import java.io.*;
import java.util.*;

public class Main {

  public static void main(String[] args) throws Exception {
    Scanner sc = new Scanner(System.in);
    PrintWriter out = new PrintWriter(System.out);

    int n = sc.nextInt();
    int m = sc.nextInt();
    ArrayList<Integer> adjList[] = new ArrayList[n];
    for (int i = 0; i < n; i++) {
      adjList[i] = new ArrayList<>();
    }

    while (m-- > 0) {
      int u = sc.nextInt() - 1, v = sc.nextInt() - 1;
      adjList[u].add(v);
      adjList[v].add(u);
    }

    ArrayList<Integer>[] groups = new ArrayList[3];
    ArrayList<ArrayList<Integer>> reps = new ArrayList<>();
    for (int i = 0; i < groups.length; i++) {
      groups[i] = new ArrayList<>();
    }

    boolean ok = true;
    for (int i = 0; i < n; i++) {
      Collections.sort(adjList[i]);
      int grpNum = 0;
      boolean found = false;
      for (ArrayList<Integer> rep : reps) {
        if (same(adjList[i], rep)) {
          groups[grpNum].add(i);
          found = true;
        }
        grpNum++;
        if (found) break;
      }
      if (!found) {
        reps.add(adjList[i]);
        if (reps.size() - 1 == 3) {
          System.out.println(-1);
          return;
        }
        groups[reps.size() - 1].add(i);
      }
    }

    ok &= reps.size() == 3;
    int grpNum = 0;
    for (ArrayList<Integer> rep : reps) {
      if (rep.size() + groups[grpNum].size() == n) {
        HashSet<Integer> set = new HashSet<>();
        for (int num : rep) set.add(num);
        for (int num : groups[grpNum]) ok &= !set.contains(num);
      } else ok = false;
      grpNum++;
    }

    if (ok) {
      int[] res = new int[n];
      grpNum = 0;
      for (ArrayList<Integer> grp : groups) {
        for (int num : grp) res[num] = grpNum;
        grpNum++;
      }

      for (int i = 0; i < n; i++) {
        if (i > 0) out.print(" ");
        out.print(res[i] + 1);
      }
      out.println();
    } else {
      out.println(-1);
    }
    out.flush();
  }

  private static boolean same(ArrayList<Integer> adjList, ArrayList<Integer> rep) {
    if (adjList.size() != rep.size()) return false;
    for (int i = 0; i < adjList.size(); i++) {
      if (adjList.get(i).intValue() != rep.get(i).intValue())
        return false;
    }
    return true;
  }

  static class Scanner {
    StringTokenizer st;
    BufferedReader br;

    public Scanner(InputStream s) {
      br = new BufferedReader(new InputStreamReader(s));
    }

    public Scanner(FileReader f) {
      br = new BufferedReader(f);
    }

    public String next() throws IOException {
      while (st == null || !st.hasMoreTokens())
        st = new StringTokenizer(br.readLine());
      return st.nextToken();
    }

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

    public long nextLong() throws IOException {
      return Long.parseLong(next());
    }

    public String nextLine() throws IOException {
      return br.readLine();
    }

    public double nextDouble() throws IOException {
      String x = next();
      StringBuilder sb = new StringBuilder("0");
      double res = 0, f = 1;
      boolean dec = false, neg = false;
      int start = 0;
      if (x.charAt(0) == '-') {
        neg = true;
        start++;
      }
      for (int i = start; i < x.length(); i++)
        if (x.charAt(i) == '.') {
          res = Long.parseLong(sb.toString());
          sb = new StringBuilder("0");
          dec = true;
        } else {
          sb.append(x.charAt(i));
          if (dec)
            f *= 10;
        }
      res += Long.parseLong(sb.toString()) / f;
      return res * (neg ? -1 : 1);
    }

    public boolean ready() throws IOException {
      return br.ready();
    }

  }
}