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.awt.*;
import java.io.*;
import java.lang.reflect.Array;
import java.math.BigInteger;
import java.util.*;

import static java.lang.Math.*;

public class Main {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    //BufferedReader br = new BufferedReader(new FileReader("hall.in"));

    StringTokenizer st = new StringTokenizer("");
    PrintWriter pw = new PrintWriter(System.out);
    //PrintWriter pw = new PrintWriter("hall.out");

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

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

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

    double nextDouble() throws IOException {
        return Double.parseDouble(next());
    }

    public Main() throws FileNotFoundException {
    }

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

    long mod = (long) 1e9 + 7;
    ArrayList<Integer>[] graph;
    long[] col;
    long[][] edges;
    long[] count;
    boolean[] was;

    void dfs(int v) {
        was[v] = true;
        for (Integer e : graph[v]) {
            if (col[v] == col[e]) {
                pw.println(-1);
                pw.close();
                System.exit(0);
            }
            if (!was[e] && col[e] == -1) {
                if (col[v] == 1 || col[v] == 3) col[e] = 2;
                else col[e] = 3;
                dfs(e);
            }
        }
    }

    void countedges(int v) {
        was[v] = true;
        for (Integer e : graph[v]) {
            edges[(int)col[v]][(int)col[e]]++;
            if (!was[e]) countedges(e);
        }
    }

    void run() throws IOException {
        int n = nextInt();
        int m = nextInt();
        count = new long[4];
        col = new long[n];
        was = new boolean[n];
        edges = new long[4][4];
        graph = new ArrayList[n];
        for (int i = 0; i < graph.length; i++) {
            graph[i] = new ArrayList<>();
        }
        for (int i = 0; i < m; i++) {
            int from = nextInt() - 1;
            int to = nextInt() - 1;
            graph[from].add(to);
            graph[to].add(from);
        }
        col[0] = 1;
        for (Integer c : graph[0]) {
            col[c] = -1;
        }
        for (int i = 0; i < col.length; i++) {
            if (col[i] == -1) continue;
            col[i] = 1;
        }
        for (int i = 0; i < was.length; i++) {
            if (!was[i]) {
                dfs(i);
            }
        }
        for (int i = 0; i < col.length; i++) {
            count[(int) col[i]]++;
        }
        was = new boolean[n];
        for (int i = 0; i < was.length; i++) {
            if (!was[i]) countedges(i);
        }
        for (int i = 1; i < count.length; i++) {
            if (count[i] == 0) {
                pw.print(-1);
                pw.close();
                return;
            }
        }
        long sum = edges[1][2] + edges[2][3] + edges[1][3];
        if (edges[1][2] != count[1] * count[2] || edges[1][3] != count[1] * count[3] || edges[2][3] != count[2] * count[3] || sum != m)
            pw.print(-1);
        else {
            for (int i = 0; i < col.length; i++) {
                pw.print(col[i] + " ");
            }
        }
        pw.close();
    }

    class Pair {
        int x, y;

        public Pair(int a, int b) {
            x = a;
            y = b;
        }
    }

    class PairComp implements Comparator<Pair> {

        @Override
        public int compare(Pair o1, Pair o2) {
            if (o1.x == o2.x) return Integer.compare(o1.y, o2.y);
            return Integer.compare(o1.x, o2.x);
        }
    }
}