// Original C++ solution (slower):
//#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();

    }

}


// 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: