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>
#define all(a) (a).begin(),(a).end()
using namespace std;

const int N = 2e5+1;
vector<int>graph[N];
int in_degree[N];
int ans[N];

void dfs(int cur,int parent=-1){
	for(int child : graph[cur]){
		ans[child] = max(ans[child],ans[cur]+1);
		in_degree[child]--;
		if(in_degree[child] == 0) dfs(child,cur);
	}
}

int main(){
	ios_base::sync_with_stdio(false);

	int n , m;
	cin >> n >> m;

	for(int i=0;i<m;i++){
		int x,y;
		cin >> x >> y;
		graph[x].push_back(y);
		in_degree[y]++;
	}

	// 0 vertex will connect with every other vertex

	for(int i=1;i<=n;i++){
		graph[0].push_back(i);
		in_degree[i]++;
	}

	dfs(0);

	cout << *(max_element(ans,ans+n+1)) - 1 << endl;

	return 0;
}