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.util.LinkedList;
import java.util.List;
import java.util.Scanner;

/**
 * 思路：
 * 1、所有图的问题，基本都是往拓扑排序上考虑
 * 2、本题通过拓扑排序判断是否存在有向环
 * 3、如果无环，所有边都输出1
 * 4、如果有环，小点到大点的边输出1，大点到小点的边输出2
 */

public class CF1217D {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();

        int[] vertices = new int[n + 1];
        int[][] edges = new int[m + 1][2];
        for(int i = 1; i <= m; i++){
            edges[i][0] = sc.nextInt();
            edges[i][1] = sc.nextInt();
            vertices[edges[i][1]]++;
        }
        if(hasCycle(vertices, edges)){
            System.out.println(2);
            for(int j = 1; j < edges.length; j++){
                if(edges[j][0] <= edges[j][1]){
                    System.out.print("1 ");
                }else{
                    System.out.print("2 ");
                }
            }
        }else{
            System.out.println(1);
            for(int j = 1; j < edges.length; j++){
                System.out.print("1 ");
            }
        }
    }

    private static boolean hasCycle(int[] vertices, int[][] edges){

        List<Integer> queue = new LinkedList<>();
        for(int i = 1; i < vertices.length; i++){
            if(vertices[i] == 0){
                queue.add(i);
            }
        }

        while(!queue.isEmpty()){
            int i = queue.remove(0);
            for(int j = 1; j < edges.length; j++){
                if(edges[j][0] == i){
                    vertices[edges[j][1]]--;
                    if(vertices[edges[j][1]] == 0){
                        queue.add(edges[j][1]);
                    }
                }
            }
        }

        for(int i = 1; i < vertices.length; i++){
            if(vertices[i] > 0){
                return true;
            }
        }
        return false;
    }

}