// Slower C++ solution:
//#pragma GCC target ("avx2,sse2")

//#pragma GCC optimization ("Ofast")

//#pragma GCC optimization ("unroll-loops")

#include <bits/stdc++.h>

#include <ext/pb_ds/assoc_container.hpp>

#include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;



#define ordered_set tree <pair <ll, int>, null_type, less <pair <ll, int> >, rb_tree_tag, tree_order_statistics_node_update>

#define ll long long

#define ull unsigned long long

#define db long double

#define pb push_back

#define pf push_front

#define ppb pop_back

#define ppf pop_front

#define fi first

#define se second

#define mp make_pair

#define up_b upper_bound

#define low_b lower_bound

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

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

#define endl "\n"



#define left fsdsdfoisf

#define sum dpsdfioppsf

#define assign xcvjlkdjfio

#define trie fksdfkjkfnjuiv

#define next sidlfjsfkl

#define merge sdfksdkfsldf



//#define int long long



using namespace std;



void dout() {

    cerr << endl;

}

template <typename Head, typename... Tail>

void dout(Head H, Tail... T) {

    cerr << H << ' ';

    dout(T...);

}



//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

typedef pair <int, int> pii;



const int N = 1e5 + 7;

int n, m, dp[N];

vector <int> q[N], r[N], ord;

bool used[N];



void dfs(int v) {

    used[v] = true;

    for (auto i : q[v]) {

        if (!used[i]) {

            dfs(i);

        }

    }

    ord.pb(v);

}



void solve(int tc) {

    // check for (int i = 0; i < n; j++)

    cin >> n >> m;

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

        int x, y;

        cin >> x >> y;

        q[x].pb(y);

        r[y].pb(x);

    }

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

        if (!used[i]) {

            dfs(i);

        }

    }

    reverse(all(ord));

    int res = 0;

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

        int v = ord[i - 1];

        for (auto j : r[v]) {

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

        }

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

    }

    cout << res;

}



signed main() {

    ios_base::sync_with_stdio(false);

    cin.tie(NULL);

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

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

    int tc = 1;

//    cin >> tc;

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

        solve(i);

//        cleanup();

    }

}


// Faster C++ solution for the same problem (measured speedup: 2.1398897385995697x):
#include <bits/stdc++.h>



using namespace std;



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

#define clr(v, d)       memset(v, d, sizeof(v))

#define rep(i, v)       for(int i=0; i<sz(v); ++i)

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

#define lpi(i, j, n)    for(int i=(j); i<(int)(n); ++i)

#define lpd(i, j, n)    for(int i=(j); i>=(int)(n); --i)

#define INF             1000000000





typedef long long ll;

typedef pair<int, int> ii;

typedef vector<ii> vii;

typedef vector<int> vi;



struct edge{

	int from;

	int to;

	int nxt_edge;

};



const int N = 1e5+5;



int n, e;

struct edge edges[N];

int heads[N];

int dp[N];



int solve(int cur_edge_idx){

	if(cur_edge_idx == -1)

		return 0;

// 	cout << cur_edge_idx << endl;

	int & cur = dp[cur_edge_idx];

	if(~cur)

		return cur;

	cur  = solve(edges[cur_edge_idx].nxt_edge);

	if(edges[cur_edge_idx].to > 0)

	    cur = max(cur, 1 + solve(heads[edges[cur_edge_idx].to]));

	return cur;

}



int main()

{

    cin >> n >> e;

	clr(edges, 0);

	clr(heads, -1);

	clr(dp, -1);

    lp(i, e){

    	cin >> edges[i].from >> edges[i].to;

    	edges[i].nxt_edge = heads[edges[i].from];

    	heads[edges[i].from] = i;

    }

    int res = 0;

    lpi(i, 1, n+1)

    	res = max(res, solve(heads[i]));

    cout << res << endl;

    return 0;

}