// Slower C++ solution:


/**

    And in the end, the love you take is equal to the love you make.



        I LOVE YOU LAURA.

**/



#include <bits/stdc++.h>

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

#include <ext/rope>



#define _CRT_SECURE_NO_DEPRECATE

#define _CRT_DISABLE_PERFCRIT_LOCKS



#define findOrder find_by_order

#define orderKey order_of_key



#define isOn(mask, i) (mask & (1 << i))

#define turn(mask, i) (mask |= (1 << i))

#define switch(mask, i) (mask &= ~(1 << i))

#define change(mask, i) (mask ^= (1 << i))

#define lowbit(mask) (mask & (-mask))

#define turnAll(mask, n) (mask = (1 << n) - 1)

#define switchLastBit(mask) (mask & (mask - 1))

#define turnLastZero(mask) (mask | (mask + 1))

#define modulo(x, mod) ((x) & (mod - 1))

#define log2(x) (31 - __builtin_clz(x))



#define mkp make_pair

#define pb push_back

#define pf push_front

#define eb emplace_back

#define ef emplace_front

#define ins insert

#define rsz resize

#define f first

#define s second

#define lb lower_bound

#define ub upper_bound



#define sz(x) ((int)x.size())

#define all(x) begin(x), end(x)

#define rall(x) rbegin(x), rend(x)



#define debug(x) (cerr << x << " ")



using namespace std;

using namespace __gnu_pbds;

using namespace __gnu_cxx;



typedef double db;

typedef long long ll;

typedef long double ld;

typedef unsigned long long ull;



typedef pair<int, int> ii;

typedef pair<ll,ll> pll;

typedef pair<ld,ld> pdb;

typedef pair<int, ii> tri;

typedef complex<ld> cd;



typedef vector<int> vi;

typedef vector<ii> vii;

typedef vector<ld> vdb;

typedef vector<cd> vcd;

typedef vector<ll> vll;

typedef vector<string> vs;

typedef vector<vll> matrix;



typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> indexed_tree;



const ll mod = 1e9+7;

const int maxn = 1e5+2;

const int pi = 2 * acos(0.0);

const db eps = 1e-9;



ll n, m;

vll graph[maxn];

ll dp[maxn];



ll dfs (int node){

    ll &ans = dp[node];



    if (ans != -1) return ans;



    for (auto v: graph[node]){

        ans = max(ans, dfs(v) + 1);

    }

    return ans;

}



int main()

{

    ios_base::sync_with_stdio(0);

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



    memset(dp, -1, sizeof dp);



    cin >> n >> m;



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

        int a, b; cin >> a >> b;



        graph[a].pb(b);

    }



    ll ans = 0;

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

        ans = max(ans, dfs(i) + 1);

    }



    cout << ans << "\n";



    return 0;

}


// Faster C++ solution for the same problem (measured speedup: 1.249600275890558x):
/*

    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;

}