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>
using namespace std; 

vector<int> FindSource(vector<int> indegree, int n)
{
    vector<int> res;
    for(int i=1; i<=n; ++i)
    {
        if (indegree[i] == 0)
            res.push_back(i);
    }

    return res;
}


int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
 
    int N, M, u, v;
    cin >> N >> M;

    vector<int> adjList[N + 1];
    vector<int> indegree(N + 1, 0);
    vector<int> distance(N + 1, 0);
    
    for(int i=0; i<M; ++i)
    {
        cin >> u >> v;
        adjList[u].push_back(v);
        ++indegree[v];
    }

    vector<int> sources = FindSource(indegree, N);

    queue<int> q;
    
    for(int source: sources)
        q.push(source);

    int ans = 0;
    while (!q.empty())
    {
        int currentNode = q.front(); q.pop();

        for(int i=0; i < adjList[currentNode].size(); ++i)
        {
            int nextNode = adjList[currentNode][i];
            distance[nextNode] = max(distance[nextNode], distance[currentNode] + 1);
            ans = max(ans, distance[nextNode]);
            --indegree[nextNode];
            if (indegree[nextNode] == 0)
            {
                q.push(nextNode);
            }
        }
    }

    cout << ans << "\n";

    return 0;
}