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 Java7 (OpenJDK 1.7.0) (judge verdict: Accepted)

/*package whatever //do not write package name here */

import java.util.*;

public class Main {
    static LinkedList<Integer>[] adj;
    static int[] dist;
    static boolean[] marked;
    static int[] indeg;
	public static void main (String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt(), m = sc.nextInt();
		adj = new LinkedList[n];
		dist = new int[n];
		indeg = new int[n];
		marked = new boolean[n];
		for(int i=0;i<n;i++){
		    adj[i] = new LinkedList<>();
		}
		for(int i=0;i<m;i++){
		    int u = sc.nextInt()-1, v = sc.nextInt()-1;
		    adj[u].add(v);
		    indeg[v]++;
		}
		for(int i=0;i<n;i++){
		    if(!marked[i] && indeg[i]==0){
		        dfs(i);
		    }
		}
		int max = 0;
		for(int i=0;i<n;i++){
		    max = Math.max(max, dist[i]);
		}
		System.out.println(max);
	}
	static void dfs(int u){
	    marked[u] = true;
	    for(int i : adj[u]){
	        dist[i] = Math.max(dist[i], dist[u]+1);
	        indeg[i]--;
	        if(indeg[i]==0) dfs(i);
	    }
	}
}