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.util.*;
import java.math.*;

public class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int N = sc.nextInt();
    int M = sc.nextInt();
    int[] x = new int[M];
    int[] y = new int[M];
    for (int i = 0; i < M; i++) {
      x[i] = sc.nextInt();
      y[i] = sc.nextInt();
    }
    sc.close();

    boolean[][] edge = new boolean[N + 1][N + 1];
    for (int i = 0; i < M; i++) edge[x[i]][y[i]] = true;

    int[] memo = new int[N + 1];
    Arrays.fill(memo, -1);

    int ans = 0;
    for (int xi = 1; xi < N + 1; xi++)
      ans = Math.max(ans, count(edge, memo, xi));

    System.out.println(ans);
  }

  private static int count(boolean[][] edge, int[] memo, int xi) {
    if (memo[xi] != -1) return memo[xi];
    int count = 0;
    boolean[] dest = edge[xi];
    for (int yi = 1; yi < dest.length; yi++)
      if (dest[yi])
        count = Math.max(count, (memo[yi] != -1 ? memo[yi] : count(edge, memo, yi)) + 1);
    memo[xi] = count;
    return count;
  }
}