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.util.Scanner;

import static java.lang.Math.max;

public class D {
    static int mar[];
    static int mas[][];

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int m = in.nextInt();
        mas = new int[m][3];
        mar = new int[n];
        for (int i = 0; i < m; i++) {
            int a = in.nextInt() - 1;
            int b = in.nextInt() - 1;
            mas[i][0] = a;
            mas[i][1] = b;
        }
        for (int i = 0; i < n; i++) {
            if (mar[i] == 0) dfs(i);
        }
        int mx = 0;
        for (int i = 0; i < m; i++) {
            mx = max(mx, mas[i][2]);
        }
        System.out.println(mx);
        for (int i = 0; i < m; i++) {
            System.out.print(mas[i][2] + " ");
        }
    }

    private static void dfs(int tek) {
        mar[tek] = 1;
        for (int i = 0; i < mas.length; i++) {
            if (mas[i][0] == tek) {
                int to = mas[i][1];
                if (mar[to] == 1) mas[i][2] = 2;
                else {
                    mas[i][2] = 1;
                    if (mar[to] == 0) dfs(to);
                }
            }
        }
        mar[tek] = 2;
    }

}
