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.LinkedList;
import java.util.StringTokenizer;

public class mainD {
    public static PrintWriter out = new PrintWriter(System.out);
    public static FastScanner enter = new FastScanner(System.in);
    static LinkedList<Integer>[] vert;
    static boolean[] was;
    static int ans[];
    static int n;

    public static void main(String[] args) throws IOException {
        n = enter.nextInt();
        int m = enter.nextInt();
        vert = new LinkedList[n + 1];
        was = new boolean[n + 1];
        ans = new int[n + 1];
        for (int i = 0; i < n + 1; i++) {
            vert[i] = new LinkedList<>();
        }
        for (int i = 0; i < m; i++) {
            int a = enter.nextInt();
            int b = enter.nextInt();
            vert[a].addLast(b);
            vert[b].addLast(a);
        }
        boolean flag = true;
        for (int i = 1; i < n + 1; i++) {
            if (!was[i]) flag &= solveThisComp(i);
            if (!flag) {
                out.println(-1);
                out.close();
                return;
            }
        }
        for (int i = 1; i < n + 1; i++) {
            out.println(ans[i] + 1);
        }
        out.close();

    }

    static int[] nVert = new int[3];

    private static boolean solveThisComp(int v) {
        nVert = new int[]{0, 0, 0};
        for (int son : vert[v]) {
            ans[son] = 1;
        }
        if (!dfs(v)) return false;
        nVert[0] = nVert[0] - vert[v].size();
        if (!dfs2(v, vert[v].size())) return false;
        if(vert[v].size()==0) return false;
        if (!dfs3(vert[v].peekFirst(), 1)) return false;
        if (!dfs4(v)) return false;
        dfsLast(v);
        return (long) nVert[0] * nVert[1] * nVert[2] != 0;
    }

    private static void dfsLast(int v) {
        was[v] = true;
        for (int son : vert[v]) {
            if (!was[son]) dfsLast(son);
        }
    }

    private static boolean dfs4(int v) {
        was[v] = false;
        if (nVert[0] + nVert[1] + nVert[2] - nVert[ans[v]] != vert[v].size()) return false;
        boolean flag = true;
        for (int son : vert[v]) {
            if (was[son]) flag &= dfs4(son);
        }
        return flag;
    }


    static boolean dfs3(int v, int color) {
        was[v] = true;
        boolean flag = true;
        if (ans[v]!=0){
            ans[v] = color;
            nVert[color]++;
            for (int son : vert[v]) {
                if (!was[son]) flag &= dfs3(son, 3 - color);
                else {
                    if (ans[son] == ans[v]) return false;
                }
            }
        }
        return flag;
    }

    static boolean dfs2(int v, int p) {
        was[v] = false;
        boolean flag = true;
        if (ans[v] == 0 && vert[v].size() != p) return false;
        for (int son : vert[v]) {
            if (was[son]) flag &= dfs2(son, p);
        }
        return flag;
    }

    static boolean dfs(int v) {
        was[v] = true;
        nVert[0]++;
        boolean answer = true;
        for (int son : vert[v]) {
            if (!was[son]) answer &= dfs(son);
            else {
                if (ans[son] == ans[v] && ans[son] == 0) return false;
            }
        }
        return answer;
    }

    static class FastScanner {
        BufferedReader br;
        StringTokenizer stok;

        FastScanner(InputStream is) {
            br = new BufferedReader(new InputStreamReader(is));
        }

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

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

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

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

        char nextChar() throws IOException {
            return (char) (br.read());
        }

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