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 class Simple3{ 

	static class Pair{
		int key;
		int value;
		Pair(int x,int y){
			this.key=x;
			this.value=y;
		}
		int getKey(){
			return this.key;
		}
		int getValue(){
			return this.value;
		}

	}

final static int N = 5010;
static ArrayList<Integer>[] g;
static ArrayList<Pair> edges;
static int color[];

public Simple3(){

	g=new ArrayList[N];
	for(int i=0;i<N;i++){
		g[i]=new ArrayList<Integer>();
	}
	edges=new ArrayList<Pair>();
	color=new int[N];
}

static boolean cycle(int u) {
	color[u] = 1;
	for (Integer v : g[u]) {
		if (color[v] == 1) {
			return true;
		} 
		else if ((color[v] == 0) && cycle(v)) {
			return true;
		}
	}
	color[u] = 2;
	return false;
}

 
public static void main(String args[]) {
	int n, m;
	Simple3 simple3=new Simple3();
	Scanner sc=new Scanner(System.in);
	n=sc.nextInt();
	m=sc.nextInt();

	for (int i = 0; i < m; ++i) {
		int u, v;
		u=sc.nextInt();
		v=sc.nextInt();
		
		g[u].add(v);
		edges.add(new Pair(u,v));
	}
	boolean dag = true;
	for (int i = 1; i <= n; ++i) {
		if ((color[i]==0) && cycle(i)) {
			dag = false;
			break;
		}
	}
	if (dag) {
		System.out.println("1");
		for (int i = 1; i <= m; ++i) {
			System.out.print("1"+" ");
		}
	} else {
		System.out.println("2");
		for (Pair e : edges) {
			if (e.getKey() < e.getValue()) {
				System.out.print("1"+" ");
			} else {
				System.out.print("2"+" ");
			}
		}
	}
}
}