Problem:
Score : 100 points

Problem Statement

There is a directed graph G with N vertices and M edges.
The vertices are numbered 1, 2, \ldots, N, and for each i (1 \leq i \leq M), the i-th directed edge goes from Vertex x_i to y_i.
G does not contain directed cycles.

Find the length of the longest directed path in G.
Here, the length of a directed path is the number of edges in it.

Constraints

All values in input are integers.

2 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq x_i, y_i \leq N

All pairs (x_i, y_i) are distinct.

G does not contain directed cycles.

Input

Input is given from Standard Input in the following format:

N M
x_1 y_1
x_2 y_2
:
x_M y_M

Output

Print the length of the longest directed path in G.

Sample Input 1

4 5
1 2
1 3
3 2
2 4
3 4

Sample Output 1

3

The red directed path in the following figure is the longest:

Sample Input 2

6 3
2 3
4 5
5 6

Sample Output 2

2

The red directed path in the following figure is the longest:

Sample Input 3

5 8
5 3
2 3
2 4
5 2
5 1
1 4
4 3
1 3

Sample Output 3

3

The red directed path in the following figure is one of the longest:

Solution in Java (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;
 
public class Main {

	public static void main(String[] args) {
    	System.out.println(metodo());   
    }
 
    public static int metodo() {
    	Scanner sc = new Scanner(System.in);
    	int n = sc.nextInt();
    	int m = sc.nextInt();
    	List<Integer>[] numbers = new ArrayList[n];
    	int[] array1 = new int[n];
        for (int i = 0; i < n; i++) {
        	numbers[i] = new ArrayList<>();
        }
        for (int  i= 0; i < m; i++) {
            int s = sc.nextInt() - 1;
            int t = sc.nextInt() - 1;
            numbers[s].add(t);
        }
        Arrays.fill(array1, -1);
        int n1 = 0;
        for (int i = 0; i < n; i++) {
            n1 = Math.max(n1, metodo2(i,array1, numbers));
        }
        return n1;
    }
 
    public static int metodo2(int v, int[] array1, List<Integer>[] numbers) {
        if (array1[v] != -1) return array1[v];
 
        int n1 = 0;
 
        for (int u: numbers[v]) {
            if (array1[u] == -1) {
            	metodo2(u,array1,numbers);
            }
            n1 = Math.max(n1, array1[u] + 1);
        }
 
        return array1[v] = n1;
    }
}