You are given a directed graph with n vertices and m directed edges without self-loops or multiple edges.

Let's denote the k-coloring of a digraph as following: you color each edge in one of k colors. The k-coloring is good if and only if there no cycle formed by edges of same color.

Find a good k-coloring of given digraph with minimum possible k.

Input

The first line contains two integers n and m (2 ≤ n ≤ 5000, 1 ≤ m ≤ 5000) — the number of vertices and edges in the digraph, respectively.

Next m lines contain description of edges — one per line. Each edge is a pair of integers u and v (1 ≤ u, v ≤ n, u ≠ v) — there is directed edge from u to v in the graph.

It is guaranteed that each ordered pair (u, v) appears in the list of edges at most once.

Output

In the first line print single integer k — the number of used colors in a good k-coloring of given graph.

In the second line print m integers c_1, c_2, ..., c_m (1 ≤ c_i ≤ k), where c_i is a color of the i-th edge (in order as they are given in the input).

If there are multiple answers print any of them (you still have to minimize k).

Examples

Input


4 5
1 2
1 3
3 4
2 4
1 4


Output


1
1 1 1 1 1 


Input


3 3
1 2
2 3
3 1


Output


2
1 1 2 


import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.Queue;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader sc = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Task solver = new Task();
        solver.solve(1, sc, out);
        out.close();
    }

    static class Task {
    	public boolean topSort(ArrayList<Integer>[] edge,int[] d) {
    		int cnt=0;
    		Queue<Integer> que=new LinkedList<Integer>();
    		for(int i=1;i<d.length;i++)
    			if(d[i]==0)
    				que.add(i);
    		cnt=que.size();
    		while(!que.isEmpty()) {
    			int u=que.poll();
    			for(int v:edge[u]) {
    				if(--d[v]==0) {
    					que.offer(v);
    					cnt++;
    				}
    			}
    		}
    		
    		
    		return  cnt==edge.length-1;
    	}
    	
        public void solve(int testNumber, InputReader sc, PrintWriter out) {
        	int n=sc.nextInt();
        	int m=sc.nextInt();
        	ArrayList<Integer>[] edge=new ArrayList[n+1];
        	int[] d=new int[n+1];
        	
        	for(int i=1;i<=n;i++)
        		edge[i]=new ArrayList<Integer>();
        	int[] x=new int[m];
        	int[] y=new int[m];
        	for(int i=0;i<m;i++) {
        		int u=sc.nextInt();
        		int v=sc.nextInt();
        		x[i]=u;
        		y[i]=v;
        		edge[u].add(v);
        		d[v]++;
        	}
        	if(topSort(edge,d)) {
        		out.println("1");
        		for(int i=1;i<=m;i++)
        			out.print("1 ");
        		out.println();
        		return ;
        	}
        	out.println("2");
        	for(int i=0;i<m;i++) {
        		if(x[i]<y[i])
        			out.print("1 ");
        		else
        			out.print("2 ");
        	}
        	out.println();
        }
    }

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

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

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

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

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

    }
}
