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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.StringTokenizer;
import java.util.Vector;
 
public class D {
    static ArrayList<int[]>[] graph;;
    static int[] col;
    static int[] res;
    static boolean isCycle;
    public static void main(String[] args) {
        FastScanner in = new FastScanner();
        int n = in.nextInt();
        int m = in.nextInt();
 
        graph=(ArrayList<int[]>[])new ArrayList[n+5];
 
        for (int i = 0; i < n; i++) {
            graph[i] = new ArrayList<>();
        }
 
        col = new int[n+5];
        res = new int[m+5];
        isCycle = false;
 
        for (int i = 0; i < m; i++) {
            int from = in.nextInt();
            int to = in.nextInt();
 
            from--; to--;
 
            graph[from].add(new int[] {to, i});
        }
 
        for (int i = 0; i < n; i++) {
            if(col[i] == 0) {
                dfs(i);
            }
        }
 
        System.out.println(isCycle ? 2 : 1);
        for (int i = 0; i < m; i++) {
            System.out.print(res[i] + " ");
        }
    }
 
    private static void dfs(int u) {
        col[u] = 1;
 
        for (int[] p : graph[u]) {
            int to = p[0];
            int id = p[1];
 
            if(col[to] == 0) {
                dfs(to);
                res[id] = 1;
            }
 
            if(col[to] == 2) {
                res[id] = 1;
            }
            else{
                res[id] = 2;
                isCycle = true;
            }
        }
        col[u] = 2;
    }
 
    static class FastScanner
    {
        BufferedReader br;
        StringTokenizer st;
 
        public FastScanner()
        {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }
 
        String next()
        {
            while (st == null || !st.hasMoreElements())
            {
                try
                {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException e)
                {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
 
        int nextInt()
        {
            return Integer.parseInt(next());
        }
 
        long nextLong()
        {
            return Long.parseLong(next());
        }
 
        double nextDouble()
        {
            return Double.parseDouble(next());
        }
 
        String nextLine()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }
            catch (IOException e)
            {
                e.printStackTrace();
            }
            return str;
        }
    }
}