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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>

#define N_MAX 100002

using namespace std;

int n, m;

vector <int> edges[N_MAX];

int indeg[N_MAX];

queue <int> q;

int d[N_MAX];

int main()
{
    cin >> n >> m;
    for(int i = 1; i <= m; i++)
    {
        int u, v;
        cin >> u >> v;
        indeg[v]++;
        edges[u].push_back(v);
    }
    for(int i = 1; i <= n; i++)
        if(indeg[i] == 0)
        {
            d[i] = 0;
            q.push(i);
        }
    int ans = 0;
    while(!q.empty())
    {
        int f = q.front();
        q.pop();
        for(int u : edges[f])
        {
            indeg[u]--;
            if(indeg[u] == 0)
            {
                d[u] = d[f] + 1;
                ans = max(ans, d[u]);
                q.push(u);
            }
        }
    }
    cout << ans << "\n";
    return 0;
}