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: Accepted)

#include<bits/stdc++.h>
using namespace std;
/*
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds ;

template <typename PB>
using ordered_set = tree<PB,null_type,less<PB>,rb_tree_tag,tree_order_statistics_node_update> ;
*/

#define     Boost               ios::sync_with_stdio(false),cin.tie(NULL),cout.tie(NULL)
#define     all(cont)           cont.begin(),cont.end()
#define     rall(cont)          cont.rbegin(),cont.rend()
#define     foreach(cont)       for(auto &val: cont)
#define     dbg(args...)        do {   cerr << #args << "-> " ;  show(args); } while(0); cerr<< endl ;
#define     TS                  template < typename T >
#define     TP                  template < typename F, typename S >
#define     TM                  template<typename T1, typename... T2>
#define     flush               cin.ignore(numeric_limits<streamsize>::max(), '\n')
#define     memz(cont)          memset(cont, 0, sizeof cont)
#define     memn(cont)          memset(cont, -1, sizeof cont)
#define     Case                printf("Case %d: ",++cs)
#define     PI                  3.1415926535897932384
#define     pf                  printf
#define     sf                  scanf
#define     mp                  make_pair
#define     pb                  push_back
#define     ff                  first
#define     ss                  second

typedef     long long                   ll ;
typedef     unsigned long long          ull ;
typedef     pair <int, int>             pii ;
typedef     pair <ll , ll >             pll ;

/** Debugging Tool **/
TS void show(const T& v) {    cerr << v << ' ' ;}
TM void show(const T1& first,const T2&... rest){show(first);show(rest...);}
TP ostream& operator<<(ostream& os,const pair<F,S>& p){return os<<"("<<p.ff<<", "<<p.ss<<")";}
TS ostream& operator<<(ostream& os,const vector<T>& v){os << "{"; typename vector< T >::const_iterator it;
    for(it=v.begin();it!=v.end();it++){if( it != v.begin() )os<<", ";os<<*it;}return os<<"}";}
TS ostream& operator<<(ostream& os,const set<T>& v){os<<"[";typename set<T>::const_iterator it;
    for(it=v.begin();it!=v.end();it++){if(it!=v.begin())os<<", ";os<<*it;}return os<<"]";}
TP ostream& operator<<(ostream& os,const map<F,S>& v){os<<"[";typename map<F,S>::const_iterator it;
    for(it=v.begin();it!=v.end();it++){if(it!=v.begin())os<<", ";os<<it->ff<<" = "<<it->ss;}return os<<"]";}
/** Ends **/

const double eps = 1e-9 ;
/*
int const dr[] = { 0, 0, 1, -1, -1, -1, 1, 1 } ;
int const dc[] = { 1, -1, 0, 0, -1, 1, -1, 1 } ;
int const drH[] = { 2, 1, -1, -2, -2, -1, 1, 2 } ;
int const dcH[] = { 1, 2, 2, 1, -1, -2, -2, -1 } ;
*/

const int lim = 1e5+5 ;

vector<int> adj[lim] ;
bool vis[lim] ;
int dp[lim] ;

int dfs(int u){
    if(vis[u]) return dp[u] ;
    vis[u] = 1 ;
    for(int v: adj[u]){
        dp[u] = max(dp[u], dfs(v)+1) ;
    }

    return dp[u] ;
}

int solve(){
    int n,m ;
    cin>> n >> m ;
    for(int i=0; i<m; i++){
        int u,v ;
        cin>> u >> v ;
        adj[u].pb(v) ;
    }

    for(int i=1; i<=n; i++)
        dfs(i) ;

    int ans = 0 ;
    for(int i=1; i<=n; i++)
        ans = max(ans, dp[i]) ;

    cout<< ans << "\n" ;


    return 0 ;
}


int main()
{
    //Boost ;
    int t = 1,cs = 0 ;
    //sf("%d",&t) ;

    while(t--){
        //Case ;
        solve() ;
    }

    return 0;
}