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: Accepted)


import java.util.*;

public class Main {
    
    long INF = Long.MAX_VALUE;
    long[] dp;
    int N;
    ArrayList<ArrayList<Integer>> G;
    int MAX_M = 100000;
    int MAX_N = 100000;
    
    public static void main(String[] args) {
        new Main().solve();
    }
    
    void solve() {
        Scanner sc = new Scanner(System.in);
        N = sc.nextInt();
        int M = sc.nextInt();
        dp = new long[N + 1];
        Arrays.fill(dp, -1);
        G = new ArrayList<ArrayList<Integer>>(N + 1);
        for (int i = 0; i <= N; i++) {
            G.add(i, new ArrayList<>());
        }
        for (int i = 0; i < M; i++) {
            int x = sc.nextInt();
            int y = sc.nextInt();
            G.get(x).add(y);
        }
        long res = 0;
        for(int v = 1;v <= N;v++){
            res = Math.max(res, rec(v));
        }
        System.out.println(res);
    }

    // dp[i]: iを始点としたときの最大の最大値
    long rec(int i) {
        long res = 0;
        if (dp[i] != -1) {
            return dp[i];
        }
        for (int j : G.get(i)) {
            res = Math.max(res, rec(j) + 1);
        }
        return dp[i] = res;
    }
}