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>
	using namespace std;

	int main() {
	#ifdef LOCAL_DEFINE
		// freopen("data.in", "rt", stdin);
		// freopen("data.out", "w", stdout);
		auto _start = chrono::high_resolution_clock::now();
	#endif

		ios::sync_with_stdio(false);
		cin.tie(0);
		cout << fixed << setprecision(15);
		int n, m;
		cin >> n >> m;
		vector<vector<int>> g(n);
		for(int i = 0; i < m; i++) {
			int x, y;
			cin >> x >> y;
			x--; y--;
			g[x].push_back(y);
		}
		vector<int> dp(n), vis(n);
		vector<int> topo;
		function<void(int)> dfs = [&](int u) {
			vis[u] = 1;
			for(int v : g[u]) {
				if(vis[v]) continue;
				dfs(v);
			}
			topo.push_back(u);
			vis[u] = 2;
		};
		for(int i = 0; i < n; i++) {
			if(vis[i]) continue;
			dfs(i);
		}
		reverse(topo.begin(), topo.end());
		int ans = 0;
		for(int i = 0; i < n; i++) {
			int u = topo[i];
			for(int v : g[u]) {
				dp[v] = max(dp[v], dp[u] + 1);
			}
		}
		for(int i = 0; i < n; i++) {
			ans = max(ans, dp[i]);
		}
		cout << ans << '\n';

	#ifdef LOCAL_DEFINE
		auto _end = chrono::high_resolution_clock::now();
		cerr << "elapsed time: " << chrono::duration<double, milli>(_end - _start).count() << " ms\n";
	#endif
		return 0;
	}