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 11.0.6) (judge verdict: Accepted)

import java.util.*;

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

        List<Integer>[] incoming = new List[n];
        int[] outcoming = new int[n];
        for (int i = 0; i < n; i++) incoming[i] = new LinkedList<>();

        for (int i = 0; i < m; i++) {
            int u = scanner.nextInt()-1;
            int v = scanner.nextInt()-1;
            outcoming[u]++;
            incoming[v].add(u);
        }

        int[] dp = new int[n];
        Queue<Integer> queue = new LinkedList<>();
        for (int i = 0; i < n; i++) {
            if(outcoming[i]==0) queue.add(i);
        }

        while (!queue.isEmpty()){
            Integer poll = queue.poll();
            int curVal = dp[poll];
            for (Integer neighbor : incoming[poll]) {
                outcoming[neighbor]--;
                dp[neighbor] = Math.max(dp[neighbor], curVal+1);
                if(outcoming[neighbor]==0) queue.add(neighbor);
                
            }
        }
        int ans = 0;
        for (int a : dp) ans = Math.max(ans, a);
        System.out.println(ans);
    }
}