stablegradients/pie-gem5-bysrc huggingface dataset:
// Original C++ solution (slower):
// Author : Syed Mujtaba

// Seek knowledge from the Cradle to the Grave !!

// https://www.linkedin.com/in/syed-mujtaba-064a5016a 



#include <bits/stdc++.h>

using namespace std;

#define int     long long

#define fast_io ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);

#define mod ((int)(1e9 + 7))

vector < int > to[100005];

vector < int > from[100005];

vector < int > roots;

int dp[100005];

int cal(int i)

{

	if(dp[i] != -1)

		return dp[i];

	int dist = -1;

	for(int j = 0; j < to[i].size(); j++)

	{

		dist = max(dist, 1 + cal(to[i][j]));

	}

	return dp[i] = dist;

}

signed main()

{

	fast_io

	int n, m;

	int x, y;

	cin >> n >> m;

	for(int i = 0; i < m; i++)

	{	

		cin >> x >> y;

		to[y].push_back(x);

		from[x].push_back(y);

	}

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

		dp[i] = -1;

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

	{

		if(to[i].size() == 0)

		{

			roots.push_back(i);

			dp[i] = 0;

		}

	}

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

	{

		if(dp[i] == -1)

			dp[i] = cal(i);

	}

	int ans = -1;

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

	{

		ans = max(ans, dp[i]);

	}

	cout << ans << endl;

}

// Optimized solution to the same problem (faster):
#include <iostream>

#include <string>

#include<iomanip>

#include <algorithm>

#include <map>

#include <vector>

#include<queue>

#include<math.h>

#include<cmath>

#include<sstream>

#include<set>

#include <map>

#include<string.h>

#include<sstream>

#include<stack>

#include<unordered_map>

#include <iostream>

#include <stdio.h>

#include <ctype.h>

#include <cstdio>

#include <cstdlib>

#define ll long long

#define ios ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

#define arr2dof vector  vector<vector<int>>dp(N,vector<int>(2, inf));

using namespace std;

const int N = 2e5 + 5, inf = 1e9+7 ;

const double ep = 0.0000000001;

const double pi = 3.14159265359;

const long long inff= 1e18;

vector < vector<int>>graph;

int n, m,a,b;

int dp[100005];

bool visited[100005];

void dfs(int node) {

	visited[node] = true;

	for (int child : graph[node]) {

		if (!visited[child]) 

		dfs(child);

		

		

			dp[node] = max(dp[node], 1 + dp[child]);

		

	}

}





int main()

{

	

	cin >> n>>m;

	graph.resize(max(n,m) + 5);

	

	for (int i = 0; i < m; ++i) {

		cin >> a >> b;

		graph[a].push_back(b);

	

	}

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

		if (!visited[i])

			dfs(i);

	}

	int ans = 0;

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

		ans = max(ans, dp[i]);

	}

	cout << ans;

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

#define eps 1e-3

#define pi acos(-1.0)

#define inf 0x3f3f3f3f

#define debug1 cout<<"&&";

#define debug2 cout<<"**";

#define pb push_back

#define ms(a, x) memset(a, x, sizeof(a))

#define forn(i, n) for(int i = 0; i < n; ++i)

#define for1(i, n) for(int i = 1; i <= n; ++i) 

typedef double db;

typedef long long ll;

typedef unsigned long long ull;

using namespace std;

const int mod = 1e9;

const int N = 1e5+10;

/*=======================================================================*/



int n, m;

int dp[N], in[N];

vector<int>G[N];

queue<int>q;

int ans = 0;



int main() {

   int a, b;

   scanf("%d%d", &n, &m);

   forn(i, m) {

      scanf("%d%d", &a, &b);

      in[b]++;

      G[a].push_back(b);

   }

   for1(i, n) {

      if(!in[i]) q.push(i);

   }

   while(!q.empty()) {

      int now = q.front();

      q.pop();

      int sz = G[now].size();

      forn(i, sz) {

         dp[G[now][i]] = max(dp[G[now][i]], dp[now] + 1);

         in[G[now][i]]--;

         if(in[G[now][i]] == 0) q.push(G[now][i]);

      }

   }

   for1(i, n) {

      ans = max(ans, dp[i]);

   }

   cout << ans << endl;

}

// Optimized solution to the same problem (faster):
#include <iostream>

#include <string>

#include<iomanip>

#include <algorithm>

#include <map>

#include <vector>

#include<queue>

#include<math.h>

#include<cmath>

#include<sstream>

#include<set>

#include <map>

#include<string.h>

#include<sstream>

#include<stack>

#include<unordered_map>

#include <iostream>

#include <stdio.h>

#include <ctype.h>

#include <cstdio>

#include <cstdlib>

#define ll long long

#define ios ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

#define arr2dof vector  vector<vector<int>>dp(N,vector<int>(2, inf));

using namespace std;

const int N = 2e5 + 5, inf = 1e9+7 ;

const double ep = 0.0000000001;

const double pi = 3.14159265359;

const long long inff= 1e18;

vector < vector<int>>graph;

int n, m,a,b;

int dp[100005];

bool visited[100005];

void dfs(int node) {

	visited[node] = true;

	for (int child : graph[node]) {

		if (!visited[child]) 

		dfs(child);

		

		

			dp[node] = max(dp[node], 1 + dp[child]);

		

	}

}





int main()

{

	

	cin >> n>>m;

	graph.resize(max(n,m) + 5);

	

	for (int i = 0; i < m; ++i) {

		cin >> a >> b;

		graph[a].push_back(b);

	

	}

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

		if (!visited[i])

			dfs(i);

	}

	int ans = 0;

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

		ans = max(ans, dp[i]);

	}

	cout << ans;

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

const int N=100005;

int dp[N],vis[N];

vector<int>e[N],p;

void dfs(int v)

{

	vis[v]=1;

	for(int i=0;i<e[v].size();i++)	

		if(!vis[e[v][i]])

			dfs(e[v][i]);

	p.push_back(v);

}

int main()

{

	int n,m;

	scanf("%d%d",&n,&m);

	for(int i=1;i<=m;i++)

	{

		int x,y;

		scanf("%d%d",&x,&y);

		e[x].push_back(y);

	}

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

		if(!vis[i])

			dfs(i);

	reverse(p.begin(),p.end());

	int ans=0;

	for(int i=0;i<n;i++)

	{

		int v=p[i];

		for(int j=0;j<e[v].size();j++)

			dp[e[v][j]]=max(dp[e[v][j]],dp[p[i]]+1);

		ans=max(ans,dp[v]);

	}

	printf("%d\n",ans);

	return 0;

}


// Optimized solution to the same problem (faster):
#include <iostream>

#include <string>

#include<iomanip>

#include <algorithm>

#include <map>

#include <vector>

#include<queue>

#include<math.h>

#include<cmath>

#include<sstream>

#include<set>

#include <map>

#include<string.h>

#include<sstream>

#include<stack>

#include<unordered_map>

#include <iostream>

#include <stdio.h>

#include <ctype.h>

#include <cstdio>

#include <cstdlib>

#define ll long long

#define ios ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

#define arr2dof vector  vector<vector<int>>dp(N,vector<int>(2, inf));

using namespace std;

const int N = 2e5 + 5, inf = 1e9+7 ;

const double ep = 0.0000000001;

const double pi = 3.14159265359;

const long long inff= 1e18;

vector < vector<int>>graph;

int n, m,a,b;

int dp[100005];

bool visited[100005];

void dfs(int node) {

	visited[node] = true;

	for (int child : graph[node]) {

		if (!visited[child]) 

		dfs(child);

		

		

			dp[node] = max(dp[node], 1 + dp[child]);

		

	}

}





int main()

{

	

	cin >> n>>m;

	graph.resize(max(n,m) + 5);

	

	for (int i = 0; i < m; ++i) {

		cin >> a >> b;

		graph[a].push_back(b);

	

	}

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

		if (!visited[i])

			dfs(i);

	}

	int ans = 0;

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

		ans = max(ans, dp[i]);

	}

	cout << ans;

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;



typedef long long ll;

typedef long double ld;

typedef pair<int,int> pii;

typedef pair<ll,ll> pll;



const int mx=1e5+5;



int dp[mx];

vector<int>g[mx];



int path(int u)

{

	if(dp[u]) return dp[u];

	if(g[u].size()==0) return dp[u]=0;

	for(auto it: g[u])

	{

		dp[u]=max(dp[u],1+path(it));

	}

	return dp[u];

}		



void solve()

{

	int n,m;

	cin>>n>>m;

	

	for(int i=1;i<=m;i++)

	{

		int u,v;

		cin>>u>>v;

		g[u].push_back(v);

	}

	

	int ans=0;

	

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

	{

		ans=max(ans,path(i));

	}

	cout<<ans;

}



int main()

{

	ios::sync_with_stdio(false);

	cin.tie(0); cout.tie(0);

	

	solve();

							                                				

	cerr<<"\nTime elapsed: " << 1000.0 * clock() / CLOCKS_PER_SEC << " ms\n";

	

	return 0;

}

// Optimized solution to the same problem (faster):
#include <iostream>

#include <string>

#include<iomanip>

#include <algorithm>

#include <map>

#include <vector>

#include<queue>

#include<math.h>

#include<cmath>

#include<sstream>

#include<set>

#include <map>

#include<string.h>

#include<sstream>

#include<stack>

#include<unordered_map>

#include <iostream>

#include <stdio.h>

#include <ctype.h>

#include <cstdio>

#include <cstdlib>

#define ll long long

#define ios ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

#define arr2dof vector  vector<vector<int>>dp(N,vector<int>(2, inf));

using namespace std;

const int N = 2e5 + 5, inf = 1e9+7 ;

const double ep = 0.0000000001;

const double pi = 3.14159265359;

const long long inff= 1e18;

vector < vector<int>>graph;

int n, m,a,b;

int dp[100005];

bool visited[100005];

void dfs(int node) {

	visited[node] = true;

	for (int child : graph[node]) {

		if (!visited[child]) 

		dfs(child);

		

		

			dp[node] = max(dp[node], 1 + dp[child]);

		

	}

}





int main()

{

	

	cin >> n>>m;

	graph.resize(max(n,m) + 5);

	

	for (int i = 0; i < m; ++i) {

		cin >> a >> b;

		graph[a].push_back(b);

	

	}

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

		if (!visited[i])

			dfs(i);

	}

	int ans = 0;

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

		ans = max(ans, dp[i]);

	}

	cout << ans;

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

vector<int>adj[100005];

int dp[100005];

bool vis[100005];

void dfs(int u)

{

    vis[u]=true;

    for(auto v:adj[u])

    {

        if(vis[v]==false)

            dfs(v);

        dp[u]=max(dp[u],dp[v]+1);

    }





}

int main()

{

    int n,m;

    cin>>n>>m;

    for(int i=1; i<=m; i++)

    {

        int u,v;

        cin>>u>>v;

        adj[u].push_back(v);

    }

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

    {

        if(vis[i]==false)

            dfs(i);

    }

    int ans=0;

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

    cout<<ans<<endl;

}


// Optimized solution to the same problem (faster):
#include <iostream>

#include <string>

#include<iomanip>

#include <algorithm>

#include <map>

#include <vector>

#include<queue>

#include<math.h>

#include<cmath>

#include<sstream>

#include<set>

#include <map>

#include<string.h>

#include<sstream>

#include<stack>

#include<unordered_map>

#include <iostream>

#include <stdio.h>

#include <ctype.h>

#include <cstdio>

#include <cstdlib>

#define ll long long

#define ios ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

#define arr2dof vector  vector<vector<int>>dp(N,vector<int>(2, inf));

using namespace std;

const int N = 2e5 + 5, inf = 1e9+7 ;

const double ep = 0.0000000001;

const double pi = 3.14159265359;

const long long inff= 1e18;

vector < vector<int>>graph;

int n, m,a,b;

int dp[100005];

bool visited[100005];

void dfs(int node) {

	visited[node] = true;

	for (int child : graph[node]) {

		if (!visited[child]) 

		dfs(child);

		

		

			dp[node] = max(dp[node], 1 + dp[child]);

		

	}

}





int main()

{

	

	cin >> n>>m;

	graph.resize(max(n,m) + 5);

	

	for (int i = 0; i < m; ++i) {

		cin >> a >> b;

		graph[a].push_back(b);

	

	}

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

		if (!visited[i])

			dfs(i);

	}

	int ans = 0;

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

		ans = max(ans, dp[i]);

	}

	cout << ans;

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;

#define ll long long

#define ull unsigned long long

#define PII pair<int,int>

#define PSI pair<string,int>

#define PIII pair< int, pair<int,int> >

const int MaxN = 1e5+5;

vector<int> g[MaxN];

int dp[MaxN];

int N;



int dfs(int node, int pr){



    auto &ret = dp[node];

    if(ret!=-1) return ret;

    ret = 0;

    for(auto i: g[node]){

        if(i==pr) continue;

        ret = max( ret, dfs(i,node)+1 );

    }



    return ret;

}



void read(){



    int i,edge;

    memset(dp,-1,sizeof dp);

    cin>>N>>edge;

    for(i=0; i<edge; i++){

        int u,v;

        cin>>u>>v;

        g[u].push_back(v);

    }

}



void solve(){



    read();

    int res = 0;

    for(int i=1; i<=N; i++){

        res = max( res, dfs(i,0) );

    }



    cout<< res << "\n";

}



int main(){



    //freopen("input.txt","r",stdin);

    ios_base::sync_with_stdio(false);

    cin.tie(nullptr);

    solve();

    return 0;

}


// Optimized solution to the same problem (faster):
#include<bits/stdc++.h>

#define si(a) scanf("%d", &a)

#define fore(i,a,b) for(int i = a; i < b; ++i)

using namespace std;

typedef long long ll;

typedef pair<int,int> pi;

typedef vector<int> vi;

vector<vector<int> > gr;

vi out;

vi ans;

vector<bool> visited;

vi aux;

void dfs(int u){

	visited[u] = true;

	//ans[u] = max(ans[u], val);

	fore(j,0,gr[u].size()){

		if(!visited[ gr[u][j] ])

			dfs(gr[u][j]);

		//ans[u] = max(ans[u], ans[ gr[u][j] ] + 1);

	}

	aux.push_back(u);

}

void solve(){

	int n,m;

	si(n); si(m);

	int x,y;

	gr.assign(n+1,vector<int>());

	out.assign(n+1,0);

	ans.assign(n+1,0);

	vector<pi> inv(m);

	aux.reserve(n);

	vi in(n+1);

	visited.assign(n+1,false);

	fore(i,0,m){

		si(x);si(y);

		inv[i] = pi(x,y);

		out[x]++;

		in[y]++;

	}

	fore(i,1,n+1)

		gr[i].reserve( in[i] );

	fore(i,0,m){

		x = inv[i].first;

		y = inv[i].second;

		gr[x].push_back(y);

	}

		

	fore(i,1,n+1)

		if(!visited[i])

			dfs(i);

	vi dp(n+1);

	//printf("aux: ");for(int i:aux) printf("%d ", i);printf("\n");

	for(int v: aux){

		if(gr[v].size() == 0)

			dp[v] = 0;

		else{

			int k = 0;

			for(int u: gr[v])

				k = max(k, dp[u]);

			dp[v] = k+1;

		}

	}

	int anss = 0;

	//printf("dp: ");for(int i:dp) printf("%d ", i);printf("\n");

	

	for(int i: dp)

		anss = max(anss, i);

	

	printf("%d\n", anss);

}

int main(){

	solve();

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;

const int maxn=1e5+5;

int dp[maxn],deg[maxn];

vector<int> G[maxn];



int get(int u)

{

	if(dp[u]) return dp[u];

	for(auto it:G[u])

	dp[u]=max(dp[u],get(it)+1);

	return dp[u];

}



int main()

{

	int n,m,x,y;

	cin>>n>>m;

	

	while(m--) 

	{

		 cin>>x>>y;

		 deg[y]++;

		 G[x].push_back(y);

	}

	

	int ans=0;

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

	if(deg[i]==0) ans=max(ans,get(i));

	

	cout<<ans<<endl;

	

	return 0;

 } 

// Optimized solution to the same problem (faster):
#include<bits/stdc++.h>

using namespace std;

#define pb push_back

#define POP pop_back()

#define ll long long

#define db double

#define POP pop_back()

#define endl '\n'

const int maxn = 1e5 + 5;

const int inf = 0x3f3f3f3f;

const int mod = 1e9 + 7;

struct edg{

	int to , nxt;

};

struct node{

	int e , s;

};

edg e[maxn];

int head[maxn] , in[maxn] , s[maxn];

int main(){

	int n , m;cin >> n >> m;

	int u , v;

	for(int i = 1;i <= m;i++){

		scanf("%d%d",&u,&v);

		e[i].nxt = head[u];

		e[i].to = v;

		head[u] = i;

		in[v]++;

	}

	queue<node>q;

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

		if(in[i] == 0)q.push((node){i,0});

	}

	node now;

	while(!q.empty()){

		now = q.front() , q.pop();

		for(int i = head[now.e];i != 0;i = e[i].nxt){

			in[e[i].to]--;

			s[e[i].to] = max(s[e[i].to] , now.s+1);

			if(in[e[i].to] == 0)q.push((node){e[i].to , s[e[i].to]});

		}

	}

	int ans = 0;

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

	cout << ans << endl;

	return 0;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <bits/stdc++.h>

using namespace std;

typedef pair<int,int> pii;

typedef vector<int> vi; 

typedef long long ll;

#define pb push_back

#define fi first

#define se second

#define rep(i,a,b) for(int i=(a);i<=(int)(b);i++)

#define per(i,a,b) for(int i=(a);i>=(int)(b);i--)

const int mod = 1e9+7;

const int maxn = 1e5+10;

const int inf = 0x3f3f3f3f;

vi g[maxn];

queue<int> q;

int dp[maxn],du[maxn];

int main()

{

    ios::sync_with_stdio(false);cin.tie(nullptr);

    //freopen("in1.txt","r",stdin);

    int n,m;

    cin>>n>>m;

    rep(i,1,m) {

        int u,v;

        cin>>u>>v;

        g[u].pb(v);

        du[v]++;

    }

    rep(i,1,n) {

        if(du[i]==0) q.push(i);

    }

    while(!q.empty()) {

        int now=q.front();q.pop();

        for(int x:g[now]) {

            du[x]--;

            if(du[x]==0) q.push(x);

            dp[x]=max(dp[x],dp[now]+1);

        }

    }

    int Max=0;

    rep(i,1,n) {

        Max=max(Max,dp[i]);

    }

    cout<<Max<<'\n';

    return 0;

}


// Optimized solution to the same problem (faster):
#include<bits/stdc++.h>

#define ll          long long int

#define ld          long double

#define pb          push_back

#define mp          make_pair

#define ff          first

#define se          second

#define scan(n)     scanf("%lld", &n);

#define print(n)    printf("%lld\n", n);

#define fd(n)       fixed<<setprecision(n)

#define MAX         1000000000000000000

#define MAXN        100005

#define LOG         19

#define mod         1000000007

using namespace std;





int main(){

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

    ll n,m,i,x,y,mx=0;

    cin>>n>>m;

    vector<ll> vi[n+1];

    ll deg[n+1];

    queue<ll> qe;

    vector<ll> vi1;

    memset(deg,0,sizeof(deg));

    for(i=1;i<=m;i++){

        cin>>x>>y;

        vi[x].pb(y);

        deg[y]++;

    }

    for(i=1;i<=n;i++){

        if(deg[i]==0){

            qe.push(i);

            vi1.pb(i);

        }

    }

    while(!qe.empty()){

        ll node=qe.front();

        qe.pop();

        for(auto x:vi[node]){

            deg[x]--;

            if(deg[x]==0){

                vi1.pb(x);

                qe.push(x);

            }

        }

    }

    ll dp[n+1];

    memset(dp,0,sizeof(dp));

    for(auto node:vi1){

        for(auto x:vi[node]){

            dp[x]=max(dp[x],dp[node]+1);

        }

    }

    for(i=1;i<=n;i++){

        mx=max(mx,dp[i]);

    }

    cout<<mx;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include<bits/stdc++.h>

#define f(i,s,v) for(int i=(int)s;i<(int)v;++i)

#define rf(i,s,v) for(int i=(int)s-1;i>(int)v;--i)

#define pb push_back

#define all(v) ((v.begin()),(v.end()))

#define memset(c, value) memset(c, (int)value, sizeof(c))

using namespace std;

typedef long long ll;

const ll oo = 1e18;

const ll MOD =1e9+7;

const double pi = acos(-1);

void fast(){

    ios_base :: sync_with_stdio(0);

    cin.tie(0);

    cout.tie(0);

}

int n, m;

bool valid (int r,int c){

    return (r < n && r >= 0 && c >= 0 && c < m);

}



int row[] = {0 , 1 , 0, -1};

int column[] = {1 , 0, -1, 0};



vector<int> adj[100001];

int path[100001];

int vis[100001];



int dp(int u){

    if(path[u]!=-1)

        return path[u];



    vis[u] = true;



    int sz = adj[u].size();



    path[u] = 0;

    f(j, 0, sz){

        if(!vis[adj[u][j]])

            path[u] = max( path[u], 1 + dp(adj[u][j]));

    }

    vis[u] = false;



    return path[u];

}





int main()

{

    fast();



    int u, v;

    cin>>n>>m;

    f(i, 0, m){

        cin>>u>>v;

        adj[u].push_back(v);

    }

    int maxi = 0;

    memset(vis, 0);

    memset(path, -1);



    f(i, 1, n+1){



        maxi = max(maxi, dp(i));

    }



    cout<<maxi<<'\n';

    return 0;

}


// Optimized solution to the same problem (faster):
#include<bits/stdc++.h>

#define ll          long long int

#define ld          long double

#define pb          push_back

#define mp          make_pair

#define ff          first

#define se          second

#define scan(n)     scanf("%lld", &n);

#define print(n)    printf("%lld\n", n);

#define fd(n)       fixed<<setprecision(n)

#define MAX         1000000000000000000

#define MAXN        100005

#define LOG         19

#define mod         1000000007

using namespace std;





int main(){

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

    ll n,m,i,x,y,mx=0;

    cin>>n>>m;

    vector<ll> vi[n+1];

    ll deg[n+1];

    queue<ll> qe;

    vector<ll> vi1;

    memset(deg,0,sizeof(deg));

    for(i=1;i<=m;i++){

        cin>>x>>y;

        vi[x].pb(y);

        deg[y]++;

    }

    for(i=1;i<=n;i++){

        if(deg[i]==0){

            qe.push(i);

            vi1.pb(i);

        }

    }

    while(!qe.empty()){

        ll node=qe.front();

        qe.pop();

        for(auto x:vi[node]){

            deg[x]--;

            if(deg[x]==0){

                vi1.pb(x);

                qe.push(x);

            }

        }

    }

    ll dp[n+1];

    memset(dp,0,sizeof(dp));

    for(auto node:vi1){

        for(auto x:vi[node]){

            dp[x]=max(dp[x],dp[node]+1);

        }

    }

    for(i=1;i<=n;i++){

        mx=max(mx,dp[i]);

    }

    cout<<mx;

}

// Fastest known solution for this problem:
// Original C++ solution (slower):
#include <iostream>

#include <stdio.h>

#include <vector>

#include <cstring>

using namespace std;

int n,m;

const int N=1e5+9;

int memo[N];

vector <vector<int> >graph;

int dp(int i)

{

    if(graph[i].empty())

        return 0;

    int &ret=memo[i];

    if(~ret)

        return ret;



    for(int u:graph[i])

    {

        ret=max(ret,dp(u)+1);

    }

    return ret;

}

int main()

{

    memset(memo,-1,sizeof memo);



    scanf("%d %d",&n,&m);

    graph.resize(n+1);

    while(m--)

    {

        int u,v;

        scanf("%d %d",&u,&v);

        graph[u].push_back(v);

    }

    int ans=0;

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

        ans=max(dp(i),ans);

    printf("%d\n",ans);



    return 0;

}


// Optimized solution to the same problem (faster):
#include<bits/stdc++.h>

#define ll          long long int

#define ld          long double

#define pb          push_back

#define mp          make_pair

#define ff          first

#define se          second

#define scan(n)     scanf("%lld", &n);

#define print(n)    printf("%lld\n", n);

#define fd(n)       fixed<<setprecision(n)

#define MAX         1000000000000000000

#define MAXN        100005

#define LOG         19

#define mod         1000000007

using namespace std;





int main(){

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

    ll n,m,i,x,y,mx=0;

    cin>>n>>m;

    vector<ll> vi[n+1];

    ll deg[n+1];

    queue<ll> qe;

    vector<ll> vi1;

    memset(deg,0,sizeof(deg));

    for(i=1;i<=m;i++){

        cin>>x>>y;

        vi[x].pb(y);

        deg[y]++;

    }

    for(i=1;i<=n;i++){

        if(deg[i]==0){

            qe.push(i);

            vi1.pb(i);

        }

    }

    while(!qe.empty()){

        ll node=qe.front();

        qe.pop();

        for(auto x:vi[node]){

            deg[x]--;

            if(deg[x]==0){

                vi1.pb(x);

                qe.push(x);

            }

        }

    }

    ll dp[n+1];

    memset(dp,0,sizeof(dp));

    for(auto node:vi1){

        for(auto x:vi[node]){

            dp[x]=max(dp[x],dp[node]+1);

        }

    }

    for(i=1;i<=n;i++){

        mx=max(mx,dp[i]);

    }

    cout<<mx;

}

// Fastest known solution for this problem: