// Original C++ solution (slower):
#include <iostream>

#include <vector>

#include <cmath>

#include <set>

#include <queue>

#include <map>

#include <cstdio>

#include <cstring>

#include <string>

#include <stack>

#include <algorithm>

#include <limits.h>

#include <bits/stdc++.h>

#define fw(x) freopen("x.txt","w",stdout)

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

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

#define REP1(i,n) for(int i = 1; i<=n;i++)

#define mem(a,s) memset(a,s,sizeof a)

#define fast ios_base::sync_with_stdio(0);cin.tie(NULL)

#define pf printf

#define sc scanf

#define ll long long

#define pii pair<int,int>

#define pll pair<ll,ll>

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

#define vii vector<int>

#define vll vector<ll>

#define vss vector<string>

#define debug(x)  cout<<#x"="<<(x)<<endl

#define pb push_back





using namespace std;





// moves



//int dx[]= {0,0,1,-1};/*4 side move*/

//int dy[]= {-1,1,0,0};/*4 side move*/

//int dx[]= {1,1,0,-1,-1,-1,0,1};/*8 side move*/

//int dy[]= {0,1,1,1,0,-1,-1,-1};/*8 side move*/

//int dx[]={1,1,2,2,-1,-1,-2,-2};/*knight move*/

//int dy[]={2,-2,1,-1,2,-2,1,-1};/*knight move*/





//big_mod



//ll bigmod(ll a,ll b,ll m)

//{if(b == 0) return 1%m;ll x = bigmod(a,b/2,m);x = (x * x) % m;if(b % 2 == 1) x = (x * a)% m;return x;}

//ll BigMod(ll B,ll P,ll M){ ll R=1%M; while(P>0) {if(P%2==1){R=(R*B)%M;}P/=2;B=(B*B)%M;} return R;} /// (B^P)%M



//ll getBit(ll num, int idx) {return ((num >> idx) & 1ll) == 1ll;}

//ll setBit1(ll num, int idx) {return num or (1ll<<idx);}

//ll setBit0(ll num, int idx) {return num & ~(1ll<<idx);}

//ll flipBit(ll num, int idx) {return num ^ (1ll<<idx);}



const int mod = 1000000007;

int n, m, x, y;

vii gr[100001];

int len[100001];

int ans=0;

int dp(int cur){

  if(len[ cur ] == -1){

    int sz = gr[cur].size();

    if(sz>0){

      int mx = 0;

      for(int j=0; j<sz; j++){

        if(len[ gr[cur][j] ] == -1) {

          len[gr[cur][j]] = dp(gr[cur][j]);

          mx = max(mx, len[gr[cur][j]]);

        }

        else mx = max(mx, len[gr[cur][j]]);

      }

      len[cur] = mx+1;

      return len[cur];

    }

    else return 0;

  }

  else return len[cur];

}

int main()

{

    fast;

    mem(len, -1);

    cin>>n>>m;

    REP1(i, m){

      cin>>x>>y;

      gr[x].pb(y);

    }

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

        int sz=gr[i].size();

        if(sz>0){

          int mx=0;

          for(int j=0; j<sz; j++){

            if(len[ gr[i][j] ] == -1) {

              len[gr[i][j]] = dp(gr[i][j]);

              mx = max(mx, len[gr[i][j]]);

            }

            else mx = max(mx, len[gr[i][j]]);

          }

          len[i] = mx+1;

          ans = max(ans, mx+1);

        }

        else len[i] = 0;

    }

    cout<<ans<<'\n';

    return 0;

}


// Optimized solution to the same problem (faster):
#include<bits/stdc++.h>

#define REP(x,y,z) for(int x=y;x<=z;x++)

#define MSET(x,y) memset(x,y,sizeof(x))

#define M 100005

using namespace std;

int n,m,dp[M];

bool vis[M];

vector<int> e[M];

int dfs(int cur) {

    if (vis[cur]) {

        return dp[cur];

    }



    int &res = dp[cur];

    for (int i:e[cur]) res = max(res, 1+dfs(i));

    vis[cur] = true;

    return res;

}

int main()

{

    scanf("%d %d", &n, &m);

    REP(i,1,m) {

        int x,y;

        scanf("%d %d", &x, &y);

        e[x].push_back(y);

    }



    int ans = 0;

    REP(i,1,n) ans = max(ans, dfs(i));

    printf("%d\n", ans);

    return 0;

}


// Fastest known solution for this problem: