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.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.HashSet;
import java.util.StringTokenizer;


public class ProblemD {
	public static void main(String[] args) throws IOException {
		InputStream in = System.in;
		InputReader scan = new InputReader(in);
		int n = scan.nextInt();
		int m = scan.nextInt();
		ArrayList<HashSet<Integer>> graph = new ArrayList<>();
		for(int i=0;i<=n;i++) {
			graph.add(new HashSet<>());
		}
		for(int i=0;i<m;i++) {
			int v1 = scan.nextInt();
			int v2 = scan.nextInt();
			graph.get(v1).add(v2);
			graph.get(v2).add(v1);
		}
		int[] allottedSets = new int[n+1];
		boolean possible = true;
		HashSet<Integer> visitedNodes = new HashSet<>();
		ArrayList<HashSet<Integer>> sets = new ArrayList<>();
		for(int set=0;set<=3;set++) {
			sets.add(new HashSet<>());
		}
		loop1:for(int set=1;set<=3;set++) {
			//Find a non-visited node
			int firstNode = -1;
			for(int i=1;i<=n;i++) {
				if(!visitedNodes.contains(i)) {
					firstNode = i;
					break;
				}
			}
			if(firstNode==-1) {
				possible = false;
				break;
			}
			allottedSets[firstNode] = set;
			sets.get(set).add(firstNode);
			visitedNodes.add(firstNode);
			//Add all unconnected node to same set if not already belongs to another set
			//If any unconnected node belongs to another set answer is impossible
			for(int i=1;i<=n;i++) {
				if(!graph.get(firstNode).contains(i)) {
					if(visitedNodes.contains(i) && i!=firstNode) {
						possible = false;
						break loop1;
					}
					allottedSets[i] = set;
					sets.get(set).add(i);
					visitedNodes.add(i);
				}
			}
		}
		//Check all nodes are visited
		if(visitedNodes.size()!=n) possible = false;
		//Check set1---set2 + set1---set3 + set2---set3 = m
		int edgeCount = 0;
		loop1:for(int set1=1;set1<=3;set1++) {
			for(int set2=set1+1;set2<=3;set2++) {
				for(int node1:sets.get(set1)) {
					for(int node2:sets.get(set2)) {
						if(!graph.get(node1).contains(node2)) {
							possible = false;
							break loop1;
						}
						edgeCount++;
					}
				}
			}
		}
		if(edgeCount!=m) possible = false;
		if(possible) {
			StringBuilder ans = new StringBuilder();
			for(int i=1;i<=n;i++) {
				ans.append(allottedSets[i]).append(" ");
			}
			System.out.println(ans.toString());
		} else {
			System.out.println(-1);
		}
	}
	static class InputReader {
		BufferedReader br;
		StringTokenizer st;
		
		InputReader(InputStream in) {
			br = new BufferedReader(new InputStreamReader(in));
		}
		
		String next() {
			while (st == null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException ioe) {
					ioe.printStackTrace();
				}
			}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		
		public long nextLong() {
			return Long.parseLong(next());
		}
	}
}
