// Slower C++ solution:
#include <bits/stdc++.h>



using namespace std;



#define el				'\n'

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

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

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



double const EPS = 1e-8, PI = acos(-1);

const int N = 1e5 + 9, M = 50 + 9, OO = (int) 1e9 + 1;

const long long MOD = 1e9 + 7;



typedef long long ll;



void OUTPUT() {

	cout << fixed << setprecision(12);

	ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);

}



void INPUT() {

#ifndef ONLINE_JUDGE

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

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

#endif

	cout << fixed << setprecision(0);

//	ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);

}



int n, m, x, y, vis[N], ans;

vector<vector<int>> edges(N);



int DFS(int node) {

	if (vis[node] != -1)

		return vis[node];

	vis[node] = 0;

	for (auto child : edges[node]) {

		DFS(child);

		vis[node] = max(vis[node], vis[child] + 1);

	}

	return 0;

}



int main() {

//	INPUT();

	OUTPUT();

	clr(vis, -1);

	cin >> n >> m;

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

		cin >> x >> y;

		edges[x].push_back(y);

	}

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

		DFS(j);

	}

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

		ans = max(ans, vis[i]);

	}

	cout << ans;



	return 0;



}


// Faster C++ solution for the same problem (measured speedup: 1.130928996396244x):
/*

    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;

}