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.io.*;
import java.util.*;

public class D {
    Random random = new Random(751454315315L + System.currentTimeMillis());
    ArrayList<Edge>[] g;
    int[] col;
    int[] ans;
    boolean one;

    class Edge {
        int to;
        int id;

        public Edge(int to, int id) {
            this.to = to;
            this.id = id;
        }
    }

    void dfs(int v) {
        col[v] = 1;
        for (Edge e : g[v]) {
            if (col[e.to] == 0) {
                ans[e.id] = 1;
                dfs(e.to);
            } else if (col[e.to] == 2) {
                ans[e.id] = 1;
            } else {
                ans[e.id] = 2;
                one = false;
            }
        }
        col[v] = 2;
    }

    public void solve() throws IOException {
        int n = nextInt(), m = nextInt();
        g = new ArrayList[n];
        for (int i = 0; i < n; i++) {
            g[i] = new ArrayList<>();
        }
        for (int i = 0; i < m; i++) {
            int u = nextInt() - 1, v = nextInt() - 1;
            g[u].add(new Edge(v, i));
        }
        col = new int[n];
        ans = new int[m];
        one = true;
        for (int i = 0; i < n; i++) {
            if (col[i] == 0) {
                dfs(i);
            }
        }
        out.println(one ? 1 : 2);
        for (int i = 0; i < m; i++) {
            out.print(ans[i] + " ");
        }
    }

    public void run() {
        try {
            br = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);

            solve();

            out.close();
        } catch (IOException e) {
            e.printStackTrace();
            System.exit(1);
        }
    }

    private void shuffle(int[] s) {
        for (int i = 0; i < s.length; ++i) {
            int j = random.nextInt(i + 1);
            int t = s[i];
            s[i] = s[j];
            s[j] = t;
        }
    }

    BufferedReader br;
    StringTokenizer in;
    PrintWriter out;

    public String nextToken() throws IOException {
        while (in == null || !in.hasMoreTokens()) {
            in = new StringTokenizer(br.readLine());
        }
        return in.nextToken();
    }

    public int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    public double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }

    public long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    public int[] nextArr(int n) throws IOException {
        int[] res = new int[n];
        for (int i = 0; i < n; i++) {
            res[i] = nextInt();
        }
        return res;
    }

    public static void main(String[] args) throws IOException {
        Locale.setDefault(Locale.US);
        new D().run();
    }
}