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.*;
import java.io.*;
public class D72
{
    static ArrayList<Edge> [] adj; static int [] visited; static int [] c; static boolean two;
    public static void main(String [] args)
    {
        MyScanner sc = new MyScanner();
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
        int n = sc.nextInt(); int m = sc.nextInt();
        adj = new ArrayList[n+1];
        for (int i = 1; i <= n; i++) adj[i] = new ArrayList<>();
        c = new int[m];
        for (int i = 0; i < m; i++) {
            Edge e = new Edge(sc.nextInt(), sc.nextInt(), i);
            adj[e.a].add(e);
        }
        visited = new int[n+1]; two = false;
        for (int i = 1; i <= n; i++) {
            if (visited[i] == 0) {
                dfs(i);
            }
        }
        out.println(two ? 2 : 1);
        for (int i = 0; i < m; i++) out.print(c[i] + " ");
        out.close();
    }

    static void dfs(int cur) {
        visited[cur] = 1;
        for (Edge next: adj[cur]) {
            if (visited[next.b] == 1) {
                c[next.i] = 2;
                two = true;
            } else if (visited[next.b] == 2) {
                c[next.i] = 1;
            } else {
                c[next.i] = 1;
                dfs(next.b);
            }
        }
        visited[cur] = 2;
    }

    static class Edge {
        int a; int b; int i;
        Edge(int a, int b, int i) {
            this.a = a; this.b = b; this.i = i;
        }
    }


    //-----------MyScanner class for faster input----------
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            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;
        }



    }

}