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 Java8 (OpenJDK 1.8.0) (judge verdict: Runtime Error)

import java.io.*;
import java.util.*;

class Main {

    // Generated by 1.1.7.1 https://github.com/kyuridenamida/atcoder-tools  (tips: You use the default template now. You can remove this line by using your custom template)
    static long rst = 0;
    public static void main(String[] args) throws Exception {
        final Scanner sc = new Scanner(System.in);
        long N;
        N = sc.nextLong();
        long M;
        M = sc.nextLong();
        long[] x = new long[(int)(M)];
        long[] y = new long[(int)(M)];
        for(int i = 0 ; i < M ; i++){
                x[i] = sc.nextLong();
                y[i] = sc.nextLong();
        }
        solve(N, M, x, y);
    }

    static void solve(long N, long M, long[] x, long[] y){
        boolean[][] G = new boolean[(int)(N+1)][(int)(N+1)];
        // System.out.print(G[0].length);
        for (int i = 0; i < M; i++){
            G[(int)x[i]][(int)y[i]] = true;
        }
        boolean[] visited = new boolean[(int)(N+1)];
        long[] dp = new long[(int)(N+1)];
        for (int i = 1; i <= (int)N; i++){
            dfs(G, i, visited, dp);
        }

        for (int i = 1; i <= (int)N; i++){
            rst = Math.max(rst, dp[i]);
        }

        
        System.out.print(rst);
    }

    static void dfs(boolean[][] G, int x, boolean[] visited, long[] dp){
        if (visited[x]){
            return;
        }
        visited[x] = true;
        for (int i = 1; i < G.length; i++){
            if (G[x][i]){
                if (!visited[i]){
                    dfs(G, i, visited, dp);
                }
                dp[x] = Math.max(dp[x], dp[i] + 1);
            }
        }
    }
}