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.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.StringTokenizer;

public class Coloring_Edges {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (Exception e) {

                    e.printStackTrace();
                }

            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (Exception e) {
                e.printStackTrace();
            }
            return str;
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }
    }
    static ArrayList<ArrayList<Integer>> adj;
    static int n;
    public static void main(String[] args) {
      FastReader in = new FastReader();
       n = in.nextInt();
       adj = new ArrayList<ArrayList<Integer>>(n);
      int m = in.nextInt();
      for(int i = 0; i < n; i++){
          adj.add(new ArrayList<>());
      }
      Edge arr [] = new Edge[m];
      for(int i = 0; i < m; i++){
          int u  = in.nextInt();
          int v =  in.nextInt();
          adj.get(u-1).add(v-1);
          arr[i] = new Edge(u,v);
      }

        if(!isCyclic()){
            System.out.println(1);
            for(int i = 0; i < m; i++){
                System.out.print(1+" ");
            }
            System.out.println();
        }else{
            System.out.println(2);
            for(int i = 0; i < m; i++){
                if(arr[i].ff < arr[i].ss){
                    System.out.print(1+" ");
                }else{
                    System.out.print(2+" ");
                }
            }
            System.out.println();
        }

    }

    private static boolean isCyclic()
    {

        // Mark all the vertices as not visited and
        // not part of recursion stack
        boolean[] visited = new boolean[n];
        boolean[] recStack = new boolean[n];


        // Call the recursive helper function to
        // detect cycle in different DFS trees
        for (int i = 0; i < n; i++)
            if (isCyclicUtil(i, visited, recStack))
                return true;

        return false;
    }
    private static boolean isCyclicUtil(int i, boolean[] visited,
                                 boolean[] recStack)
    {

        // Mark the current node as visited and
        // part of recursion stack
        if (recStack[i])
            return true;

        if (visited[i])
            return false;

        visited[i] = true;

        recStack[i] = true;
        List<Integer> children = adj.get(i);

        for (Integer c: children)
            if (isCyclicUtil(c, visited, recStack))
                return true;

        recStack[i] = false;

        return false;
     }


}
class Edge{
    int ff;
    int ss;

    public Edge(int ff,int ss){
        this.ff = ff;
        this.ss = ss;

    }

}