Problem: 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.

Examples

Input

4 5
1 2
1 3
3 2
2 4
3 4


Output

3


Input

6 3
2 3
4 5
5 6


Output

2


Input

5 8
5 3
2 3
2 4
5 2
5 1
1 4
4 3
1 3


Output

3

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;

class Main
{
	int V;
	ArrayList<Integer> adj[];
    Main(int V)
    {
        this.V = V;
        adj = new ArrayList[V];
        for(int i = 0; i < V; i++)
        {
            adj[i] = new ArrayList<>();
        }
    }
    void addEdge(int p1, int p2)
    {
        adj[p1].add(p2);
    }
	public static void main(String[] args) 
	{
		Scanner sc = new Scanner(System.in);
	    int n = sc.nextInt();
	    int e = sc.nextInt();
	    Main g1 = new Main(n);
	    for(int i = 0; i < e; i++)
	    {
	    	int a = sc.nextInt();
	    	int b = sc.nextInt();
	    	g1.addEdge(a-1, b-1);
	    }
	    boolean[] vis = new boolean[n];
	    int[] dp = new int[n];
	    for(int i = 0; i < n; i++)
	    {
	    	dp[i] = 0;
	    	vis[i] = false;
	    }
	    int res = 0;
	    for(int i = 0; i < n; i++)
	    {
	    	if(!vis[i])
	    	{
	    		res = Math.max(res, g1.dfs(i, dp, vis));
	    	}	
	    }
		System.out.println(res);
	}
	int dfs(int v, int[] dp, boolean[] vis)
    {
    	int ans = 0;
    	vis[v] = true;
    	Iterator<Integer> itr = adj[v].iterator();
        while(itr.hasNext())
        {
        	int i = itr.next();
        	if(!vis[i])
        	{
        		ans = Math.max(ans, dfs(i, dp, vis));
        	}	
        	else
        	{
        		ans = Math.max(ans, dp[i]);
        	}  
        }
        if(adj[v].size() == 0)
        {
        	dp[v] = ans;
        }
        else
        {
        	dp[v] = ans + 1;
        }
        return dp[v];
    }
}