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.text.*;

public class D1228 {

	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		int n = sc.nextInt();
		int m = sc.nextInt();
		HashSet<Integer> adjList[] = new HashSet[n];
		for (int i = 0; i < adjList.length; i++) {
			adjList[i] = new HashSet<Integer>();
		}
		if(m==0) {
			pw.println(-1);
			pw.close();
			return;
		}
		int u = sc.nextInt() - 1;
		int v = sc.nextInt() - 1;
		adjList[u].add(v);
		adjList[v].add(u);

		
		int[] b = new int[m];
		int[] p = new int[m];
		b[0] = u;
		p[0] = v;
		for (int i = 1; i < m; i++) {
			int x = sc.nextInt() - 1;
			int y = sc.nextInt() - 1;
			b[i] = x;
			p[i] = y;
			adjList[x].add(y);
			adjList[y].add(x);
		}
		int w = -1;
		for (int x : adjList[u]) {
			if (adjList[v].contains(x)) {
				w = x;
				break;
			}
		}
		if (w == -1) {
			pw.println(-1);
			pw.close();
			return;
		}
		int[] clr = new int[n];
		clr[u] = 1;
		clr[v] = 2;
		clr[w] = 3;
		for (int i = 0; i < n; i++) {
			if (adjList[i].contains(u) && adjList[i].contains(v) && !adjList[i].contains(w)) {
				clr[i] = 3;
			} else if (adjList[i].contains(u) && !adjList[i].contains(v) && adjList[i].contains(w)) {
				clr[i] = 2;
			} else if (!adjList[i].contains(u) && adjList[i].contains(v) && adjList[i].contains(w)) {
				clr[i] = 1;
			} else {
				pw.println(-1);
				pw.close();
				return;
			}
		}
		for (int i = 0; i < m; i++) {
			if (clr[b[i]] == clr[p[i]]) {
				pw.println(-1);
				pw.close();
				return;
			}
		}
		int[] g = new int[4], c = new int[4];
		for (int i = 0; i < n; i++) {
			g[clr[i]]++;
			c[clr[i]] += adjList[i].size();
		}
//		pw.println(Arrays.toString(clr));
//		pw.println(Arrays.toString(g));
//		pw.println(Arrays.toString(c));
		if ((c[1] == (g[2] + g[3]) * g[1]) && (c[2] == (g[1] + g[3]) * g[2]) && (c[3] == (g[2] + g[1]) * g[3])) {
			for (int i = 0; i < n; i++) {
				pw.print(clr[i] + " ");
			}
		} else {
			pw.println(-1);
		}
		pw.close();

	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream s) {
			br = new BufferedReader(new InputStreamReader(s));
		}

		public Scanner(FileReader r) {
			br = new BufferedReader(r);
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				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();
		}

		public double nextDouble() throws IOException {
			String x = next();
			StringBuilder sb = new StringBuilder("0");
			double res = 0, f = 1;
			boolean dec = false, neg = false;
			int start = 0;
			if (x.charAt(0) == '-') {
				neg = true;
				start++;
			}
			for (int i = start; i < x.length(); i++)
				if (x.charAt(i) == '.') {
					res = Long.parseLong(sb.toString());
					sb = new StringBuilder("0");
					dec = true;
				} else {
					sb.append(x.charAt(i));
					if (dec)
						f *= 10;
				}
			res += Long.parseLong(sb.toString()) / f;
			return res * (neg ? -1 : 1);
		}

		public boolean ready() throws IOException {
			return br.ready();
		}

	}
}
