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 Rust (1.42.0) (judge verdict: Accepted)

use proconio::input;
use proconio::marker::Usize1;

macro_rules! chmax {
    ($max_v: expr, $v: expr) => {
        if $v > $max_v {
            $max_v = $v;
            true
        } else {
            false
        }
    };
}

fn main() {
    input! {
        n: usize,
        m: usize,
        xy: [(Usize1, Usize1); m],
    };
    let mut e = vec![vec![]; n];
    let mut d = vec![0; n]; // 入次数 (indegree)
    for &(x_i, y_i) in xy.iter() {
        e[x_i].push(y_i);
        d[y_i] += 1;
    }

    let mut q = d
        .iter()
        .enumerate()
        .filter(|&(_, &d_i)| d_i == 0)
        .map(|(v_i, _)| v_i)
        .collect::<std::collections::VecDeque<usize>>();

    let mut dp = vec![0; n];
    while let Some(u) = q.pop_front() {
        for &v in e[u].iter() {
            chmax!(dp[v], dp[u] + 1);
            d[v] -= 1;
            if d[v] == 0 {
                q.push_back(v);
            }
        }
    }

    let ans = dp.iter().max().unwrap();
    println!("{}", ans);
}