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.util.*;
import java.io.*;
import java.math.BigInteger;
 
public class Main
{
	static final int mod = (int)1e9+7;
	static ArrayList<Integer>[] adj;
	static boolean[] vis;
	static int[] col, cnt;
	public static void main(String[] args) throws Exception
	{
		FastReader in = new FastReader();
		PrintWriter out = new PrintWriter(System.out);
		int n = in.nextInt();
		int m = in.nextInt();
		adj = new ArrayList[n + 1];
		col = new int[n + 1];
		vis = new boolean[n + 1];
		for(int i = 1; i <= n; i++)
			adj[i] = new ArrayList();
		for(int i = 0; i < m; i++)
		{
			int a = in.nextInt();
			int b = in.nextInt();
			adj[a].add(b);
			adj[b].add(a);
		}
		cnt = new int[4];
		int c = 0;
		for(int i = 1; i <= n; i++)
		{
			if(col[i] == 0)
			{
				c++;
				if(c > 3)
					break;
				for(int j = 1; j <= n; j++)
					vis[j] = false;
				dfs(i, c);
			}
		}
		if(c != 3)
		{
			System.out.print(-1);
			return;
		}
		c = cnt[1] * cnt[2] + cnt[1] * cnt[3] + cnt[2] * cnt[3];
		if(m != c)
			out.print(-1);
		else
			for(int i = 1; i <= n; i++)
				out.print(col[i]+" ");
		out.flush();
	}

	static void dfs(int ind, int c)
	{
		vis[ind] = true;

		boolean f = true;
		if(col[ind] != 0)
			f = false;

		for(int i : adj[ind])
		{
			if(col[i] == c)
				f = false;
		}
		if(f)
		{
			col[ind] = c;
			cnt[c]++;
		}

		for(int i : adj[ind])
		{
			if(!vis[i])
				dfs(i, c);
		}
	}
}
 
class FastReader
{
	BufferedReader br;
	StringTokenizer st;
 
	public FastReader()
	{
	    br = new BufferedReader(new InputStreamReader(System.in));
	}
 
	public String next() throws IOException
	{
	    if(st == null || !st.hasMoreElements())
	    {
	        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();
	}
}