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.FileReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.Queue;
import java.util.StringTokenizer;

public class D {
	static InputReader in = new InputReader("");

	public static void main(String[] args) {
		StringBuilder sb = new StringBuilder();

		n = in.nextInt();
		m = in.nextInt();
		adj = new ArrayList[n];
		int[] u = new int[m];
		int[] v = new int[m];
		for (int i = 0; i < n; i++)
			adj[i] = new ArrayList<>();
		for (int i = 0; i < m; i++) {
			u[i] = in.nextInt() - 1;
			v[i] = in.nextInt() - 1;
			adj[u[i]].add(v[i]);
			adj[v[i]].add(u[i]);
		}

		color = new int[n];
		Arrays.fill(color, 1);
		for (int i : adj[0])
			color[i] = 0;
		dfs(0);

		long one = 0, two = 0, three = 0;
		for (int i = 0; i < n; i++) {
			if (color[i] == 1)
				one++;
			if (color[i] == 2)
				two++;
			if (color[i] == 3)
				three++;
		}
		if (one * two * three == 0) {
			System.out.println(-1);
			return;
		}
		long oneTwo = one * two, oneThree = one * three, twoThree = two * three;
		for (int i = 0; i < m; i++) {
			if (color[u[i]] == color[v[i]]) {
				System.out.println(-1);
				return;
			}
			if (color[u[i]] + color[v[i]] == 3)
				oneTwo--;
			if (color[u[i]] + color[v[i]] == 4)
				oneThree--;
			if (color[u[i]] + color[v[i]] == 5)
				twoThree--;
		}
		if (oneTwo > 0 || oneThree > 0 || twoThree > 0) {
			System.out.println(-1);
			return;
		}

		for (int i = 0; i < n; i++) {
			sb.append(color[i]);
			sb.append(' ');
		}

		System.out.println(sb.toString().trim());
	}

	static int n, m;
	static ArrayList<Integer> adj[];
	static int[] color;

	static void dfs(int cur) {
		for (int i : adj[cur]) {
			if (color[i] != 0)
				continue;
			if (color[cur] == 2)
				color[i] = 3;
			else
				color[i] = 2;
			dfs(i);
		}
	}

	static class InputReader {
		private BufferedReader reader;
		private StringTokenizer tokenizer;

		public InputReader(String filename) {
			try {
				if (filename.length() == 0)
					reader = new BufferedReader(new InputStreamReader(System.in));
				else
					reader = new BufferedReader(new FileReader(filename));
			} catch (Exception e) {
			}
		}

		public String nextLine() {
			try {
				return reader.readLine();
			} catch (Exception e) {
				return tokenizer.nextToken();
			}
		}

		public String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (Exception e) {
				}
			}
			return tokenizer.nextToken();
		}

		public int nextInt() {
			return Integer.parseInt(next());
		}

		public double nextDouble() {
			return Double.parseDouble(next());
		}

		public long nextLong() {
			return Long.parseLong(next());
		}

		public int[] nextIntArray(int n) {
			int[] array = new int[n];
			for (int i = 0; i < n; i++)
				array[i] = nextInt();
			return array;
		}

		public double[] nextDoubleArray(int n) {
			double[] array = new double[n];
			for (int i = 0; i < n; i++)
				array[i] = nextDouble();
			return array;
		}

		public long[] nextLongArray(int n) {
			long[] array = new long[n];
			for (int i = 0; i < n; i++)
				array[i] = nextLong();
			return array;
		}

	}

}
