Problem: 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 

Solution (in JAVA):

Solution:
/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
public class Codechef
{
	public static void main (String[] args) throws java.lang.Exception
	{
		// your code goes here
		Scanner s = new Scanner(System.in);
		int n = s.nextInt();
		int m = s.nextInt();
		ArrayList<Pair>[] grph = new ArrayList[n];
		for(int i=0;i<n;i++)
		    grph[i] = new ArrayList<>();
		int[] ans = new int[m];
		for(int i=0;i<m;i++)
		{
		    int u = s.nextInt();
		    int v = s.nextInt();
		    grph[u-1].add(new Pair(v-1, i));
		}
		boolean[] vis = new boolean[n];
		boolean[] inpath = new boolean[n];
		boolean iscycle = false;
		for(int i=0;i<n;i++)
		{
		    if(!vis[i])
		        iscycle |= dfs(grph, i, ans, vis, inpath);
		}
		if(!iscycle)
		    System.out.println(1);
		else
		    System.out.println(2);
		for(int i=0;i<m;i++)
		    System.out.print(ans[i]+" ");
	}
	
	public static boolean dfs(ArrayList<Pair>[] grph, int u, int[] ans, boolean[] vis, boolean[] inpath)
	{
	    vis[u] = true;
	    inpath[u] = true;
	    boolean iscycle = false;
	    for(int i=0;i<grph[u].size();i++)
	    {
	        Pair p = grph[u].get(i);
	        int v = p.first;
	        if(inpath[v])
	        {
	            iscycle = true;
	            ans[p.second] = 2;
	        }
	        else if(vis[v])
	        {
	            ans[p.second] = 1;
	        }
	        else
	        {
	            ans[p.second] = 1;
	            iscycle |= dfs(grph, v, ans, vis, inpath);
	        }
	    }
	    inpath[u] = false;
	    return iscycle;
	}
	
	public static class Pair{
	    int first;
	    int second;
	    
	    Pair(int first, int second)
	    {
	        this.first = first;
	        this.second = second;
	    }
	}
	
}