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.*;

import static java.lang.Math.*;

public class D {
    static boolean cycle = false;
    static boolean vis[], inStack[];
    static ArrayDeque<Integer> g[];

    public static void main(String[] args) throws Exception {
        Scanner in = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        int n = in.nextInt(), m = in.nextInt();
        g = genDQ(n + 1);
        int edges[][] = new int[m][];
        for (int i = 0; i < m; i++) {
            int u = in.nextInt(), v = in.nextInt();
            g[u].addLast(v);
            edges[i] = new int[]{u, v};
        }

        if (!isCycle(g, n + 1)) {
            pw.println(1);
            for (int i = 1; i <= m; i++) {
                pw.print("1 ");
            }
            pw.println();
        } else {
            pw.println(2);
            for (int e[] : edges) {
                if (e[0] > e[1]) pw.print(1 + " ");
                else pw.print(2 + " ");
            }
            pw.println();
        }

        pw.close();
    }


    static boolean isCycle(ArrayDeque<Integer> graph[], int n) {
        cycle = false;
        vis = new boolean[n];
        inStack = new boolean[n];
        g = graph;
        for (int i = 0; i < n && !cycle; i++) {
            if (!vis[i]) isCycle(i);
        }
        return cycle;
    }

    static void isCycle(int u) {
        if (cycle) return;
        inStack[u] = true;
        vis[u] = true;
        for (int v : g[u]) {
            if (inStack[v]) {
                cycle = true;
                return;
            } else if (!vis[v]) isCycle(v);
        }
        inStack[u] = false;
    }


    static <T> ArrayDeque<T>[] genDQ(int n) {
        ArrayDeque<T> list[] = new ArrayDeque[n];
        for (int i = 0; i < n; i++) list[i] = new ArrayDeque<T>();
        return list;
    }

    static void debug(Object... obj) {
        System.err.println(Arrays.deepToString(obj));
    }
}