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: Wrong Answer)

#include <bits/stdc++.h>

using namespace std;
using   vi = vector<int>;
using   vvi = vector<vi>;
using   vs = vector<string>;
#define ll long long
#define mod 1000000007
#define ffor(i,a,b) for(int i=a;i<b;i++)
#define bfor(i,a,b) for(int i=a-1;i>=b;i--)
#define mp make_pair
#define pb push_back
#define ff first
#define ss second
#define mem(x,y) memset(x,y,sizeof(x))
#define all(x) x.begin(),x.end()
#define SP(x) setprecision(x)
#define sz(x) (int)x.size()
#define fast  ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
#define PI 3.14159265358979323846
#define lb lower_bound
#define ub upper_bound
#define bs binary_search
const int maxn=1e5+10;
vector<int> v[maxn];
int vis[maxn];
int a[maxn];
int solve(int k)
{
    if(vis[k]==1)
    {
        return a[k];
    }
    vis[k]=1;
    int o=0;
    for(auto x: v[k])
    {
        o=max(o,solve(x)+1);
    }
    a[k]=o;
    return o;
}
int main()
{     int n,m;
      cin>>n>>m;
      ffor(i,0,n)
      {
          int x,y;
          cin>>x>>y;
          v[x].pb(y);
      }
      int ans=0;
      ffor(i,1,n+1)
      {    
            // cout<<a[i]<<" ";
            ans=max(ans,solve(i));
      }
      cout<<ans;
	return 0;
}