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 <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;
#define SIZE_OF_ARRAY(array) (sizeof(array)/sizeof(array[0]))

static int n, m, i, j, k, x, y;
static vector<int> from[100000 + 10], to[100000 + 10];
static bool used[100000 + 10];
int dfs(int n,int cost)
{
	int ans = 0;
	//printf("dfs:now%dcost%d\n", n, cost);
	used[n]++;
	if (to[n].size() == 0) return cost;
	for (int i = 0; i < to[n].size(); i++)
	{
		ans = max(ans, dfs(to[n][i], cost + 1));
	}
	return ans;
}

int main()
{
	int ans = 0;
	queue<int> q;
	cin >> n >> m;
	for (i = 0; i < m; i++)
	{
		cin >> x >> y;
		from[y].push_back(x);
		to[x].push_back(y);
	}
	for (i = 1; i <= n; i++)
	{
		used[i] = false;
		/*
		printf("ver%dfrom:", i);
		for (j = 0; j < from[i].size(); j++)
		{
			printf(",%d", from[i][j]);
		}
		printf("ver%dto:", i);
		for (j = 0; j < to[i].size(); j++)
		{
			printf(",%d", to[i][j]);
		}
		cout << endl;
		*/
	}
	for (i = 1; i <= n; i++)
	{
		//printf("fori%db%d\n", i, (int)used[i]);
		if (used[i]) continue;
		used[i]++;
		ans=max(ans,dfs(i,0));
	}
	cout << ans << endl;
}