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: Time Limit Exceeded)

/*Author : Hrishi*/

/*When I wrote this, only God and I understood what I was doing.
Now, only God knows.*/

#include <bits/stdc++.h>

#define ll long long
#define vi vector<int>
#define vl vector<ll>
#define pii pair<int,int>
#define pll pair<ll,ll> 
#define PB push_back
#define F(i,L,R) for(auto i=L; i<R ;i++)
#define FR(i,L,R) for(auto i=R-1;i>=L;i--)
#define show(x) cout<<(#x)<<" is "<<(x)<<"\n";
#define BOOST ios_base::sync_with_stdio(false);cin.tie(NULL);
#define M 1000000007

using namespace std;

vi graph[100001];
int ans[100001];

//O(n square)

int main() 
{
    int n,m,a,b;
    cin>>n>>m;
    F(i,0,m)
    {
        cin>>a>>b;
        a--;b--;
        graph[a].PB(b);
    }
	int ans=0;
	F(i,0,n)
	{
	    vi level(n,0);
	    queue<int> q;
	    q.push(i);
	    while(!q.empty())
	    {
	        int s = q.front();
	        q.pop();
	        for(auto it=graph[s].begin();it!=graph[s].end();it++)
	        {
	            if(level[*it]<level[s]+1)
	            {
	                level[*it] = level[s] + 1;
	                q.push(*it);
	            }
	        }
	    }
	    ans = max(ans, (*max_element(level.begin(),level.end())));
	}
	cout<<ans<<"\n";
}