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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Task solver = new Task();
        solver.solve(1, in, out);
        out.close();
    }

    static class Task {
        ArrayList<ArrayList<Integer>> childs;
        boolean[] inNode;
        boolean[] outNode;
        boolean cycle;

        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int n = in.nextInt();
            int m = in.nextInt();
            childs = new ArrayList<>();
            for (int i = 0; i <= n; i++) {
                childs.add(new ArrayList<>());
            }
            int[] u = new int[m];
            int[] v = new int[m];
            for (int i = 0; i < m; i++) {
                u[i] = in.nextInt();
                v[i] = in.nextInt();
                childs.get(u[i]).add(v[i]);
            }
            inNode = new boolean[n + 1];
            outNode = new boolean[n + 1];
            cycle = false;
            for (int i = 1; i <= n; i++) {
                if (!outNode[i]) {
                    dfs(i);
                }
            }
            if (cycle) {
                out.println(2);
                for (int i = 0; i < m; i++) {
                    out.print((u[i] > v[i]) ? 1 : 2);
                    out.print(' ');
                }
            } else {
                out.println(1);
                for (int i = 1; i <= m; i++) {
                    out.print(1);
                    out.print(' ');
                }
            }
        }

        void dfs(int node) {
            if (inNode[node]) {
                if (!outNode[node]) {
                    cycle = true;
                }
            } else {
                inNode[node] = true;
                for (int child : childs.get(node)) {
                    dfs(child);
                }
                outNode[node] = true;
            }
        }

    }

    static class InputReader {
        private BufferedReader reader;
        private StringTokenizer stt;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream));
        }

        public String nextLine() {
            try {
                return reader.readLine();
            } catch (IOException e) {
                return null;
            }
        }

        public String next() {
            while (stt == null || !stt.hasMoreTokens()) {
                stt = new StringTokenizer(nextLine());
            }
            return stt.nextToken();
        }

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

    }
}