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.*;
import java.util.*;

public class D_ColoringEdges {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader inp = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Solver solver = new Solver();
        solver.solve(inp, out);
        out.close();
    }

    private static class Solver {
        private boolean cycle(ArrayList<Integer>[] graph, int root) {
            boolean[] vis = new boolean[graph.length];
            Stack<Integer> stack = new Stack<>();
            stack.add(root);
            while (!stack.isEmpty()) {
                int node = stack.pop();
                if (node == root && vis[root]) return true;
                if (vis[node]) continue;
                vis[node] = true;
                stack.addAll(graph[node]);
            }
            return false;
        }

        private void solve(InputReader inp, PrintWriter out) {
            int n = inp.nextInt(), m = inp.nextInt();
            ArrayList<Integer>[] graph = new ArrayList[n];
            int[] set = new int[m];
            for (int i = 0; i < n; i++) graph[i] = new ArrayList<>();
            for (int i = 0; i < m; i++) {
                int u = inp.nextInt() - 1, v = inp.nextInt() - 1;
                graph[u].add(v);
                set[i] = (u < v) ? 1 : 2;
            }
            boolean cycle = false;
            for (int i = 0; i < n; i++) if (cycle(graph, i)) cycle = true;
            out.println(cycle ? 2 : 1);
            for (int i = 0; i < m; i++) out.print((cycle ? set[i] : 1) + " ");
        }
    }

    static class InputReader {
        BufferedReader reader;
        StringTokenizer tokenizer;

        InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }
        public long nextLong() {
            return Long.parseLong(next());
        }
    }
}