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:
import java.util.*;
public final class code
{
	static class Pair
	{
		int first;
		int second;
		Pair(int a,int b)
		{
			first=a;
			second=b;
		}
	}
	static int n,m,res[],col[];
	static boolean cyc;
	static ArrayList<ArrayList<Pair>> list;
	static void dfs(int v)
	{
		col[v]=1;
		for(Pair p : list.get(v))
		{
			if(col[p.first]==0)
			{
				dfs(p.first);
				res[p.second]=1;
			}
			else if(col[p.first]==2)
			{
				res[p.second]=1;
			}
			else
			{
				cyc=true;
				res[p.second]=2;
			}
		}
		col[v]=2;
	}
	public static void main(String args[])
	{
		Scanner sc=new Scanner(System.in);
		n=sc.nextInt();
		m=sc.nextInt();
		list=new ArrayList<ArrayList<Pair>>();
		for(int i=0;i<n;i++)
			list.add(new ArrayList<Pair>());
		for(int i=0;i<m;i++)
		{
			int u=sc.nextInt();
			int v=sc.nextInt();
			list.get(u-1).add(new Pair(v-1,i));	
		}
		res=new int[m];
		col=new int[n];
		for(int i=0;i<n;i++)
		{
			if(col[i]==0)
				dfs(i);
		}
		System.out.println(cyc?2:1);
		for(int i=0;i<m;i++)
			System.out.print(res[i]+" ");
		System.out.println();
	}
}