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
import java.io.*;
import java.util.*;

public class Main{

    public static void main(String[] args){
		new Main().run();
    }

    void run(){
      Scanner sc=new Scanner(System.in);
      int N=sc.nextInt();
      int M=sc.nextInt();
      ArrayList<Integer>[] g=new ArrayList[N];
      for(int i=0;i<N;++i){
          g[i]=new ArrayList<>();
      }
      boolean[] valid=new boolean[N];
      Arrays.fill(valid,true);
      int[] in=new int[N];
      int[] out=new int[N];
      for(int i=0;i<M;++i){
          int x=sc.nextInt();
          int y=sc.nextInt();
          --x;
          --y;
          g[x].add(y);
          ++in[y];
          ++out[x];
      }
      ArrayDeque<Integer> pend=new ArrayDeque<>();
      for(int i=0;i<N;++i){
          if(in[i]==0)pend.add(i);
      }
      int[] dp=new int[N];
      while(!pend.isEmpty()){
          int src=pend.poll();
          for(int dst:g[src]){
            dp[dst]=Math.max(dp[dst],dp[src]+1);
            --in[dst];
            if(in[dst]==0){
                pend.add(dst);
            }
	    }
	}
	int ans=0;
	for(int v:dp)ans=Math.max(ans,v);
      System.out.println(ans);
    }

    void tr(Object...o){System.out.println(Arrays.deepToString(o));}
}
