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: Time Limit Exceeded)

#include <iostream>
#include<bits/stdc++.h>
using namespace std;

int main()
{
   long long  n,m;
   cin>>n>>m;
  if(m==1) 
  {
      cout<<m;
      return 0;
  }
   vector<vector<long long >> adj(n+1);
   long long  a,b;
   vector<long long> vec(n+1,0);
   for(int i=1;i<=m;i++)
   {
       cin>>a>>b;
       adj[a].push_back(b);
       vec[b]++;
   }
    priority_queue<pair<long long ,long long>> q;
    for(long long i=1;i<=n;i++)
    {
        if(vec[i]==0)
           q.push({0,i});
    }
    // vector<vector<long long>> vis(n+1,vector<long long>(n+1,0));
    long long ans=0;
    while(!q.empty())
    {
        pair<long long,long long> t= q.top();
        q.pop();
       if(t.first>ans)
         ans=t.first;
        for(auto y:adj[t.second])
        {
            // if(vis[t.second][y]==0)
            // {
                q.push({t.first+1,y});
            //     vis[t.second][y]=1;
            // }
        }
    }
    cout<<ans; 
    return 0;
}