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.ArrayList;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;
import java.util.StringTokenizer;

public class D {

	@SuppressWarnings("unchecked")
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		StringTokenizer st = new StringTokenizer(sc.nextLine());
		int n = Integer.parseInt(st.nextToken());
		int m = Integer.parseInt(st.nextToken());
		ArrayList<Integer>[] adj = new ArrayList[n];
		for (int i = 0; i < n; i++) {
			adj[i] = new ArrayList<Integer>();
		}
		for (int i = 0; i < m; i++) {
			st = new StringTokenizer(sc.nextLine());
			int a = Integer.parseInt(st.nextToken()) - 1;
			int b = Integer.parseInt(st.nextToken()) - 1;
			adj[a].add(b);
			adj[b].add(a);
		}
		// System.out.println(Arrays.toString(adj));

		boolean works = true;

		int[] col = new int[n]; // 0 = indeterminate, 1 = red, 2 =blue, 3 = green
		int[] depth = new int[n];
		Arrays.fill(depth, -1);
		ArrayList<Integer> bi = new ArrayList<Integer>();
		
		for(int i =0; i < n; i++) {
			if(depth[i] == -1) {
				col[i] = 1;
				depth[i] = 0;
				
				Queue<Integer> q = new LinkedList<Integer>();
				q.add(i);
				W: while (!q.isEmpty()) {
					int v = q.poll();
					if (depth[v] > 2) {
						works = false;
						break;
					}
					if (depth[v] == 0) {
						for (int w : adj[v]) {
							depth[w] = depth[v] + 1;
							q.add(w);
						}
					} else if (depth[v] == 1) {
						bi.add(v);
						for (int w : adj[v]) {
							if (col[w] == 0 && depth[w] != 1) {
								depth[w] = depth[v] + 1;
								col[w] = col[i];
								q.add(w);
							}
						}
					} else {
						for (int w : adj[v]) {
							if (col[w] == 0 && depth[w] != 1) {
								works = false;
								break W;
							}
						}
					}
				}
			}
		}
		
		//System.out.println(Arrays.toString(col));
		if (!works) {
			System.out.println(-1);
		} else {
			//System.out.println(Arrays.toString(col));
			for(int i : bi) {
				if(col[i] == 0) {
					Queue<Integer> q = new LinkedList<Integer>();
					q.add(i);
					col[i] = 2;
					W2: while (!q.isEmpty()) {
						int v = q.poll();
						
						for (int w : adj[v]) {
							
							if (depth[w] == 1) {
								if (col[w] == 0) {
									col[w] = 5 - col[v];
									q.add(w);
								} else if (col[w] == col[v]) {
									works = false;
									break W2;
								}
							}
							
						}
					}
				}
			}
			//System.out.println(Arrays.toString(col));
			if(!works) {
				System.out.println(-1);
			} else {
				int[] num = new int[4];
				for(int i =0; i <n; i++) {
					num[col[i]] ++;
				}
				if(num[0] > 0 || num[1] == 0 || num[2] == 0 || num[3] == 0) {
					works = false;
				}
				for(int i =0; i <n; i++) {
					if(col[i] == 1) {
						int ad = num[2]+num[3];
						if(adj[i].size() != ad) {
							works = false;
							break;
						}
					}
					if(col[i] == 2) {
						int ad = num[1]+num[3];
						if(adj[i].size() != ad) {
							works = false;
							break;
						}
					}
					if(col[i] == 3) {
						int ad = num[2]+num[1];
						if(adj[i].size() != ad) {
							works = false;
							break;
						}
					}
				}
				//System.out.println("still working " + works);
				if(!works) {
					System.out.println("-1");
				} else {
					for(int i =0; i < n; i++) {
						System.out.print(col[i] + " ");
					}
					System.out.println();
				}
			}
		}
		sc.close();
	}

}
