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: Accepted)

#include <bits/stdc++.h>
#pragma GCC optimize("unroll-loops,no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
typedef long long ll;
typedef long double ld;

vector <vector<ll>> graph;
stack <ll> t;
vector <bool> visited;

void tsort(ll v){
	if (visited[v]) return;
	visited[v] = true;
	for (int z=0;z<graph[v].size();z++){
		if (visited[graph[v][z]]) continue;
		tsort(graph[v][z]);
	}
	t.push(v);
}

int main(){
	ll n,m;
	cin >> n >> m;
	graph.resize(n+1);
	visited.resize(n+1,false);
	for (int z=0;z<m;z++){
		ll u,v;
		cin >> u >> v;
		graph[u].push_back(v);
	}
	for (int z=1;z<=n;z++){
		tsort(z);
	}
	vector <ll> arr(n);
	for (int z=0;z<n;z++){
		arr[z] = t.top();
		t.pop();
	}
	vector <ll> dp(n+1,0);
	for (int z=0;z<n;z++){
		for (int x=0;x<graph[arr[z]].size();x++){
			dp[graph[arr[z]][x]] = max(dp[graph[arr[z]][x]],dp[arr[z]]+1);
		}
	}
	ll ans = 0;
	for (int z=0;z<=n;z++){
		ans = max(ans,dp[z]);
	}
	cout << ans << endl;
	cin >> n;
}