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 Graph {
    
    static int V;
    static int E;
    static boolean cyc;
    static ArrayList<pair> adj[];
    static int color[];
    static int res[];
    @SuppressWarnings("unchecked")
	public Graph(int v, int e) {
        V = v;
        E = e;
        adj = new ArrayList[v];
        color = new int[v];
        res = new int[e];
        for(int i=0;i<v;i++) {
            adj[i] = new ArrayList<>();
        }
    }
    
    void addEdge(int u, int v, int i) {
        adj[u].add(new pair(v, i));
    }
    
    void isCycle(int u) {
        color[u] = 1;
        for(pair p: adj[u]) {
        	if(color[p.v]==0) {
        		isCycle(p.v);
        		res[p.i] = 1;
        	} else if(color[p.v] == 2) {
        		res[p.i] = 1;
        	} else {
        		cyc = true;
        		res[p.i] = 2;
        	}
        }
        color[u] = 2;
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int v = sc.nextInt();
        int e = sc.nextInt();
        Graph g = new Graph(v, e);
        for(int i=0;i<e;i++) {
            int src = sc.nextInt()-1;
            int dst = sc.nextInt()-1;
            g.addEdge(src, dst, i);
        }
        for(int i=0;i<v;i++) {
        	if(color[i]==0) {
        		g.isCycle(i);
        	}
        }
        System.out.println(cyc?2:1);
		for(int i=0;i<e;i++)
			System.out.print(res[i]+" ");
		System.out.println();
    }
}

class pair {
	int v, i;
	public pair(int v, int i) {
		this.v = v;
		this.i = i;
	}
}