// Slower C++ solution:
#include <iostream>

#include <stdio.h>

#include <string>

#include <cstring>

#include <cstdio>

#include <algorithm>

#include <cmath>

#include <vector>

#include <set>

#include <map>

#include <unordered_map>

#include <queue>

#include <deque>

#include <stack>

#include <bitset>

#include <iomanip>



#define ll long long

#define ld long double

#define ull unsigned long long

#define endl "\n"

#define MohamedMotaz ios::sync_with_stdio(0); cin.tie(0); ios_base::sync_with_stdio(0);

#define f(a, b, c) for(int a = b; a < c; a++)

#define fr(a, b, c) for(int a = b; a >= c; a--)

using namespace std;





//ll fact[1000001]; ll inv[1000001];

//ll primes[100007];

//ll arr[1000007];

//ll modPower(ll b, ll p) {

//	if (p == 0)

//		return 1;

//

//	ll halfpow = modPower(b, p / 2);

//	ll toReturn = (halfpow * halfpow) % mod;

//	if (p % 2)

//		toReturn = (toReturn * b) % mod;

//

//	return toReturn;

//}

//

//ll fastPower(ll b, ll p) {

//	if (p == 0)

//		return 1;

//	ll ans = fastPower(b, p / 2);

//	ans = (ans * ans);

//	if (p % 2 != 0)

//		ans = (ans * b);

//	return ans;

//}

//ll GcdRecursive(ll a, ll b) {

//	if (b == 0) return a;

//	return GcdRecursive(b, a % b);

//}

//ll modLCM(ll a, ll b) {

//	ll val = GcdRecursive(a, b);

//	ll tmp = ((a % mod) * (b % mod)) % mod;

//	ll finalVal = ((tmp % mod) * (arr[val] % mod)) % mod;

//	return finalVal;

//

//}

//ll LCM(ll a, ll b) {

//	return (a * b) / GcdRecursive(a, b);

//}

//void move1step(ll& a, ll& b, ll q) { // a and b by reference

//	ll c = a - q * b;

//	a = b;

//	b = c;

//}

//ll GcdIterative(ll a, ll b) {

//	while (b) move1step(a, b, a / b);

//	return a;

//}

//

//void pre(ll n) {

//

//	fact[0] = 1;

//	inv[0] = 1;

//

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

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

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

//		arr[i] = modPower(i, mod - 2);

//	}

//}

//

//ll npr(ll n, ll r) {

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

//}

//

//ll ncr(ll n, ll r) {

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

//}

//

//void sieve(ll val) {

//	memset(primes, 1, sizeof primes);

//	primes[0] = primes[1] = false;

//	for (int i = 2; i <= val; i++) {

//		if (primes[i]) {

//			for (int j = i * i; j <= val; j += i) {

//				primes[j] = 0;

//			}

//		}

//	}

//

//}



const ll mod = 1e9 + 7;

const ll N = 2e5 + 7;

const ll inf = 1e9 + 5;







ll n,m, a, b, ans = 0, c= 0;

vector<ll> adj[N];

bool visited[N] = {false};

ll dp[N] = {};

void dfs(ll node){





    visited[node] = true;

    for (auto elem: adj[node]){

        if (!visited[elem]) dfs(elem);

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

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

    }





}



int main()

{

    MohamedMotaz;

    cin >> n >> m;

    f(i,0,m){

        cin >> a >> b;

        adj[a].push_back(b);

    }



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



    cout << ans;

}








// Faster C++ solution for the same problem (measured speedup: 1.8330202955169528x):
//拓扑排序+dp 

#include <bits/stdc++.h>

using namespace std;

const int maxn = 100025;

int dp[maxn],in[maxn];

vector<int> G[maxn];

int main()

{

    queue<int> q;

    int n,m,a,b,ans = 0;

	cin>>n>>m;

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

    	cin>>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]);

    cout<<ans<<endl;

    return 0;

}