// Slower C++ solution:
#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;

}


// Faster C++ solution for the same problem (measured speedup: 1.2482814952633703x):
/*

    if you can't see the repay

    Why not just work step by step

    rubbish is relaxed

    to ljq

*/

#include <cstdio>

#include <cstring>

#include <iostream>

#include <queue>

#include <cmath>

#include <map>

#include <stack>

#include <set>

#include <sstream>

#include <vector>

#include <stdlib.h>

#include <algorithm>

using namespace std;



#define dbg(x) cout<<#x<<" = "<< (x)<< endl

#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl

#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl

#define max3(a,b,c) max(a,max(b,c))

#define min3(a,b,c) min(a,min(b,c))

#define lc (rt<<1)

#define rc (rt<<11)

#define mid ((l+r)>>1)



typedef pair<int,int> pll;

typedef long long ll;

const int inf = 0x3f3f3f3f;

const int _inf = 0xc0c0c0c0;

const ll INF = 0x3f3f3f3f3f3f3f3f;

const ll _INF = 0xc0c0c0c0c0c0c0c0;

const ll mod =  (int)1e9+7;



ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}

ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}

void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);

const int MAX_N = 100025;

int dp[MAX_N],in[MAX_N];

vector<int > G[MAX_N];

int main()

{

    //ios::sync_with_stdio(false);

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

    //freopen("b.txt","w",stdout);

    queue<int > q;

    int n,m,a,b,ans = 0;scanf("%d%d",&n,&m);

    for(int i = 1;i<=m;++i) scanf("%d%d",&a,&b),in[b]++,G[a].push_back(b);

    for(int i = 1;i<=n;++i) if(!in[i]) q.push(i);

    while(!q.empty())

    {

        int now = q.front();

        q.pop();

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

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

        {

            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]);

        }

    }

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

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

    //fclose(stdin);

    //fclose(stdout);

    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;

    return 0;

}