// Slower C++ solution:
#include <bits/stdc++.h>

using namespace std;

/*

#define mod 7

typedef long long ll;



ll fact[1000006], inv[1000006];



ll fp(ll base, ll exp)

{

    if (exp == 0)

        return 1;

    ll ans = fp(base, exp / 2);

    ans = (ans * ans) % mod;

    if (exp % 2 != 0)

        ans = (ans * (base % mod)) % mod;

    return ans;

}



void calcFacAndInv(ll n)

{

    fact[0] = inv[0] = 1;

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

    {

        fact[i] = (i * fact[i - 1]) % mod;

        inv[i] = fp(fact[i], mod - 2);

    }

}



ll ncr(ll n, ll r)

{

    return ((fact[n] * inv[r]) % mod * inv[n - r]) % mod;

}



ll npr(ll n, ll r)

{

    return (fact[n] * inv[n - r]) % mod;

}



int main()

{

    calcFacAndInv(100000);

     cout<<inv[3];

    return 0;

}

*/





// GCD



int gcd(long long a,long long b)

{

    if (b==0)

        return a;

    return gcd (b,a%b);

}

//prime factoriation

/*

set<int >s;

bool prime factorization (int x)

{

    for (int j=2; j*j<=x; ++j)

    {

        while (x%j==0)

        {

            s.insert(j);

            x/=j;

        }

    }

    if (x>1)

    {

        s.insert(x);

    }

    if (s.size()==2)

        return true ;

    else

        return false ;

}



#define ll long long

int a,b,q,p,n;

int countDivisorss(int x,int y)

{

    ll cnt = 0;

    for (int i = min(x,y); i <= sqrt(n); i++)

    {

        if (x % i == 0&&y%i==0)

        {

            if (x / i == i||y / i == i)

                cnt++;



            else

                cnt = cnt + 2;

        }

    }

    return cnt;

}

int countDivisors(int x)

{

    ll cnt = 0;

    int tmp=n;

    if (n-x>10000000)  tmp=sqrt(n);

    for (int i = x; i <= tmp; i++)

    {

        if (x % i == 0)

        {

         if (x/i==i)

            cnt++;

         else

            cnt = cnt + 2;

        }

    }

    return cnt;

}

*/



/*

        int s =n/2,e=n-1;

        while (s<=e&&e<=n-1)

        {

            int mid =s+e/2;



            if (arr[mid]==arr[n/2])

            {

                s=mid;

                num +=mid -s;

            }

            else

                e=mid-1;

        }

        */





/*



const int N = 1e5 + 5;



vector<int> adj[N];

bool visited[N];

int nodesNum;

void dfs(int node)

{

    if (visited[node])

        return;

    visited[node] = true;

    for (int child: adj[node])

        dfs(child);

}



//fast power



#define mod 1000000007

long long fp(long long  base, long long exp)

{

    if (exp == 0)

        return 1;

    long long  ans = fp(base, exp / 2);

    ans = (ans * ans);

    if (exp % 2 != 0)

        ans = (ans * (base )) ;

    return ans;

}



*/

#define IO   ios::sync_with_stdio(false)

/*

vector<int> v,t;

int n,V,a[21],MAX;

void dfs(int x,int sum)

{

    if(x==n)

    {

        if(sum>MAX&&sum<=V)

        {

            MAX=sum;

            t=v;

        }

        return;

    }

    v.push_back(a[x]);

    dfs(x+1,sum+a[x]);

    v.pop_back();

    dfs(x+1,sum);

}

*/

const int N=1e5+7;

int n,m,x,y;

int dp[N];

bool vis[N];

vector<int> adj[N];

void dfs(int node)

{

    vis[node] = 1;

    for (int child = 0; child< adj[node].size(); child++)

    {

        if (!vis[adj[node][child]])

            dfs(adj[node][child]);

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

    }

}

int LongestPath()

{

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

    {

        if (!vis[i])

            dfs(i);

    }

    int ans = 0;

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

    {

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

    }

    return ans;

}

int main()

{

    cin>>n>>m;

    while(m--)

    {

        cin>>x>>y;

        adj[x].push_back(y);

    }

    cout << LongestPath();

    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 2.5286701173355004x):
#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;

}