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.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.math.BigInteger;
import java.util.*;

public class Main {
	
	public static void main(String[] args) throws IOException {
		int n = in.nextInt();
		int m = in.nextInt();
		TreeSet <Integer>[] g = new TreeSet[n];
		for(int i = 0; i< n; i ++)
			g[i] = new TreeSet <> ();
		for(int i = 0; i < m; i ++) {
			int u = in.nextInt() - 1;
			int v = in.nextInt() - 1;
			g[u].add(v);
			g[v].add(u);
		}
		int[] ans = new int[n];
		ans[0] = 1;
		int next = -1;
		for(int i = 1; i < n; i ++) {
			if(g[0].contains(i)) {
				next = i;
				continue;
			}
			// i must be with 0
			ans[i] = 1;
			for(int j : g[i]) {
				if(!g[0].contains(j)) {
					System.out.println(-1);
					return;
				}
			}
			if(g[i].size() != g[0].size()) {
				System.out.println(-1);
				return;
			}
		}
		if(next == -1) {
			// 0 is alone
			System.out.println(-1);
			return;
		}
		int nextnext = -1;
		ans[next] = 2;
		for(int i = 0; i < n; i ++) {
			if(g[next].contains(i)) {
				if(ans[i] != 1)
					nextnext = i;
				continue;
			}
			// i must be with next
			ans[i] = 2;
			for(int j : g[i]) {
				if(!g[next].contains(j)) {
					System.out.println(-1);
					return;
				}
			}
			if(g[i].size() != g[next].size()) {
				System.out.println(-1);
				return;
			}
		}
		if(nextnext == -1) {
			// there is not the last group
			System.out.println(-1);
			return;
		}
		ans[nextnext] = 3;
		for(int i = 0; i < n; i ++) {
			if(g[nextnext].contains(i)) 
				continue;
			// i must be with nextnext
			ans[i] = 3;
			for(int j : g[i]) {
				if(!g[nextnext].contains(j)) {
					System.out.println(-1);
					return;
				}
			}
			if(g[i].size() != g[nextnext].size()) {
				System.out.println(-1);
				return;
			}
		}
		for(int i = 0; i < n; i ++)
			if(ans[i] == 0) {
				System.out.println(-1);
				return;
			}
		for(int i = 0; i < n; i ++)
			out.write(ans[i] + " ");
		out.flush();
	}
 
	public static FastReader in = new FastReader();
	public static BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));
}

class FastReader {
    BufferedReader br;
    StringTokenizer st;
 
    public FastReader() {
        br = new BufferedReader(new InputStreamReader(System.in));
    }
    
    public String next() {
        while (st == null || !st.hasMoreElements()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException  e) {
                e.printStackTrace();
            }
        }
        return st.nextToken();
    }
 
    public char nextChar() {
    	return next().charAt(0);
    }
    
    public int nextInt() {
        return Integer.parseInt(next());
    }
 
    public long nextLong() {
        return Long.parseLong(next());
    }
 
    public double nextDouble() {
        return Double.parseDouble(next());
    }
 
    public String nextLine() {
        String str = "";
        try {
            str = br.readLine();
        } catch (IOException e) {
            e.printStackTrace();
        }
        return str;
    }
    
    public int[] nextIntArray(int n) {
    	int[] a = new int[n];
    	for(int i = 0; i < n; i ++)
    		a[i] = nextInt();
    	return a;
    }
    
    public Integer[] nextIntegerArray(int n) {
    	Integer[] a = new Integer[n];
    	for(int i = 0; i < n; i ++)
    		a[i] = nextInt();
    	return a;
    }
    
    public long[] nextLongArray(int n) {
    	long[] a = new long[n];
    	for(int i = 0; i < n; i ++)
    		a[i] = nextLong();
    	return a;
    }
    
    public double[] nextDoubleArray(int n) {
    	double[] a = new double[n];
    	for(int i = 0; i < n; i ++)
    		a[i] = nextDouble();
    	return a;
    }
    
    public String[] nextStringArray(int n) {
    	String[] a = new String[n];
    	for(int i = 0; i < n; i ++)
    		a[i] = next();
    	return a;
    }
}