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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
#define gc getchar_unlocked
#define fo(i,n) for(i=0;i<n;i++)
#define Fo(i,k,n) for(i=k;k<n?i<n:i>n;k<n?i+=1:i-=1)
#define ll long long
#define si(x)	scanf("%d",&x)
#define sl(x)	scanf("%lld",&x)
#define ss(s)	scanf("%s",s)
#define pi(x)	printf("%d\n",x)
#define pl(x)	printf("%lld\n",x)
#define ps(s)	printf("%s\n",s)
#define deb(x) cout << #x << "=" << x << endl
#define deb2(x, y) cout << #x << "=" << x << "," << #y << "=" << y << endl
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define all(x) x.begin(), x.end()
#define clr(x) memset(x, 0, sizeof(x))
#define sortall(x) sort(all(x))
#define tr(it, a) for(auto it = a.begin(); it != a.end(); it++)
#define PI 3.1415926535897932384626
typedef pair<int, int>	pii;
typedef pair<ll, ll>	pl;
typedef vector<int>		vi;
typedef vector<ll>		vl;
typedef vector<pii>		vpii;
typedef vector<pl>		vpl;
typedef vector<vi>		vvi;
typedef vector<vl>		vvl;
int mpow(int base, int exp);
void ipgraph(int m);
//void dfs(int u, int par);
const int mod = 1000000007;
const int N = 3e5, M = N;
vector<int>adj[100000];
int dp[100000];
bool vis[10000];
int in_degree[100000];
void dfs(int src){
    vis[src]=true;    
    for(auto it:adj[src]){
        dp[it]=max(dp[it],dp[src]+1);
        --in_degree[it];
        if(in_degree[it]==0)dfs(it);
        }



}





int main(){
ios_base::sync_with_stdio(0);
ll int t,n,i,j,k,m;
t=1;
while(t--){
    cin>>n>>m;
    fo(i,m){
        int u,v;
        cin>>u>>v;
        //u--;
        //v--;
        adj[u-1].pb(v-1);
        in_degree[v-1]++;
}
memset(dp,-1,sizeof dp);
int res=0,ans=0;
fo(i,n){
//vector<int>vis(n,0);
res=0;
if(!vis[i]&&in_degree[i]==0)
dfs(i);



}
ans=*max_element(dp,dp+n);

    cout<<ans+1;
}

}