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.*;
import java.util.*;

public class Main
{
	public static ArrayList<pair> adj[];
	public static int n, m , color[],ans[];
	public static pair[] p;
	public static boolean cycle;
    public static void main(String[]fa)
    {
        Scanner scanner=new Scanner(System.in);
        n=scanner.nextInt();
        m=scanner.nextInt();
        adj=new ArrayList[n];
        for(int i=0;i<n;i++)adj[i]=new ArrayList<pair>();
        for(int i=0;i<m;i++)
        {
        	int x=scanner.nextInt()-1;
        	int y=scanner.nextInt()-1;
        	adj[x].add(new pair(y,i));
        }
        color=new int[n];
        ans=new int[m];
        cycle=false;
        for(int i=0;i<n;i++)
        if(color[i]==0)
        dfs(i);
        int num=cycle?2:1;
        System.out.println(num);
        for(int i:ans)System.out.print(i+" ");
       
    }
    
    public static  void dfs(int cur )
    {
      color[cur]=1;
      
      for(pair p:adj[cur])
      {
    	  int to=p.u;
    	  int id=p.id;
    	  if(color[to]==0)
    	  {
    		  ans[id]=1;
    		  dfs(to);
    	  }
    	  else if(color[to]==2)
    	  {
    		ans[id]=1;
    	  }
    	  else {
    		  cycle=true;
    		  ans[id]=2;
    	  }
      }
      color[cur]=2;
    }
    
    public static class pair{
    	int u ,id;
    	public pair(int u, int id)
    	{
    		this.u=u;
    		this.id=id;
    	}
    }
}