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++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

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

#define int long long
#define endl '\n'
#define ll long long
#define ld long double
#define all(x) x.begin(), x.end()
#define vi vector<int>
#define vvi vector<vi>
#define pb push_back
#define pi pair<int,int>
#define vpi vector<pi>

const int MOD = 1e9 + 7;
const ll INF = 1e18;
const int MX = 3e5 + 5;
const ld PI = 4 * atan((ld)1);
int in[MX];
vi adj[MX];
vi adj1[MX];
bool vis[MX];
int val[MX];
vi ans;
int cur=0,n,m;
void topsort(){
	while(ans.size()!=n){
		for(int i=1;i<=n;i++){
			if(in[i]==0 and !vis[i]){
				ans.pb(i);
				vis[i]=true;
			}
		}
		int i=ans[cur];
		cur++;
		for(auto it:adj[i]){
			in[it]--;
		}
	}
}
int32_t main(){
	ios_base::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	cin>>n>>m;
	for(int i=0;i<m;i++){
		int x,y;
		cin>>x>>y;
		adj[x].pb(y);
		adj1[y].pb(x);
		in[y]++;
	}
	topsort();
	for(int i=0;i<ans.size();i++){
		for(auto it:adj1[ans[i]]){
			val[ans[i]]=max(1+val[it],val[ans[i]]);
		}
	}
	cout<<*max_element(val,val+MX);
}