Problem:
Score : 100 points

Problem Statement

There is a directed graph G with N vertices and M edges.
The vertices are numbered 1, 2, \ldots, N, and for each i (1 \leq i \leq M), the i-th directed edge goes from Vertex x_i to y_i.
G does not contain directed cycles.

Find the length of the longest directed path in G.
Here, the length of a directed path is the number of edges in it.

Constraints

All values in input are integers.

2 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq x_i, y_i \leq N

All pairs (x_i, y_i) are distinct.

G does not contain directed cycles.

Input

Input is given from Standard Input in the following format:

N M
x_1 y_1
x_2 y_2
:
x_M y_M

Output

Print the length of the longest directed path in G.

Sample Input 1

4 5
1 2
1 3
3 2
2 4
3 4

Sample Output 1

3

The red directed path in the following figure is the longest:

Sample Input 2

6 3
2 3
4 5
5 6

Sample Output 2

2

The red directed path in the following figure is the longest:

Sample Input 3

5 8
5 3
2 3
2 4
5 2
5 1
1 4
4 3
1 3

Sample Output 3

3

The red directed path in the following figure is one of the longest:

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

//




#include <bits/stdc++.h>
using namespace std;


#define spdio ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define pb push_back
#define pp pop_back()
#define pf push_front
#define ppf pop_front()
#define ff first
#define ss second
#define mp make_pair
#define fr(i,j,n) for(ll i=j; i<n; i++)
#define rfr(i,j,n) for(ll i=n-1; i>=j; i--)
#define pll pair<ll,ll>
#define plb pair<ll,bool>
#define psl pair<string,ll>
#define psb pair<string,bool>
#define srt(v) sort((v).begin(),(v).end())
#define ub(v,val) upper_bound((v).begin(),(v).end(),(val))-(v).begin()
#define lb(v,val) lower_bound((v).begin(),(v).end(),(val))-(v).begin()


typedef long double ld;
typedef long long int ll;


//getline(cin,string);
const ll inf = (ll) 1e18 + 19;
const ll mod = (ll) 1e9  +  7;




ll dfs(vector <ll> v[],vector <ll> &dp,ll node) {
    ll mx=-inf;
    if(dp[node]!=-1) return dp[node];
    if(v[node].size()==0) return 0;
    for(auto it: v[node]) {
        mx=max(mx,dfs(v,dp,it));
    }
    return dp[node]=1+mx;
}




void cazz() {
    ll t=1,i=0,j,n;
    //cin>>t;
    while(t--) {
        ll m;
        cin>>n>>m;
        vector <ll> v[n+1];
        ll a,b;
        fr(i,0,m) {
            cin>>a>>b;
            v[a].pb(b);
        }
        ll mx=-inf;
        vector <ll> dp(n+1,-1);
        fr(i,1,n+1) {
            mx=max(mx,dfs(v,dp,i));
        }
        cout<<mx<<"\n";
    }
}




int main() {
    spdio;
	cazz();
	return 0;
}