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.awt.*;
import java.math.BigInteger;
import java.util.*;

public class Ada{
    static LinkedList<Integer> adj[];
    static boolean[] vis,stack;
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n=sc.nextInt(),m=sc.nextInt();
        adj=new LinkedList[n];
        for (int i=0;i<n;i++)adj[i]=new LinkedList<>();
        int color[]=new int[m];
        for (int i=0;i<m;i++){
            int u=sc.nextInt()-1,v=sc.nextInt()-1;
            adj[u].add(v);
            color[i]=u<v?1:2;
        }
        vis=new boolean[n];
        stack=new boolean[n];
        int k=1;
        for (int i=0;i<n;i++){
            if (vis[i])continue;
            if (dfs(i)){k=2;break;}
        }
        if (k==1){
            Arrays.fill(color,1);
        }
        System.out.println(k);
        StringBuilder sb=new StringBuilder();
        for (int i=0;i<m;i++){
            sb.append(color[i]+" ");
        }
        System.out.println(sb);

    }
    static boolean dfs(int s){
        if (stack[s])return true;
        if (vis[s])return false;
        vis[s]=stack[s]=true;
        for (int i:adj[s]){
            if (dfs(i))return true;
        }
        stack[s]=false;
        return false;
    }
}