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 C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
//#define int long long
#define FAST_IO ios_base::sync_with_stdio(false), cin.tie(nullptr)
using namespace std;

vector<vector<int>> adj;

vector<bool> visited;
vector<int> dist;

void dfs(int i)
{
    visited[i] = true;
    for (int v : adj[i])
    {
        if (!visited[v])
        {
            dfs(v);
        }
        dist[i] = max(dist[i], 1 + dist[v]);
    }
}
void init(int n)
{
    adj.resize(n, vector<int>());
    visited.resize(n);
    dist.resize(n);
}

signed main()
{
    FAST_IO;

    int N, M;
    cin >> N >> M;
    init(N + 1);
    while (M--)
    {
        int x, y;
        cin >> x >> y;
        adj[x].push_back(y);
    }
    for (int i = 1; i < N + 1; i++)
    {
        if (!visited[i])
        {
            dfs(i);
        }
    }

    cout << *max_element(dist.begin(), dist.end());
    return 0;
}