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.util.*;
import java.io.*;
public class Main{
    public  static int cal(Integer i,int[]visit,int[]dp,HashMap<Integer,ArrayList<Integer>>reversed_edges){
        if(visit[i]==1)return dp[i];
        else{
            dp[i]=0;
            visit[i]=1;
            if(reversed_edges.containsKey(i)){
                for(Integer m:reversed_edges.get(i)){
                dp[i]=Math.max(dp[i],cal(m,visit,dp,reversed_edges)+1);
                }
            }
        }
        return dp[i];
    }
    public static void main(String[] args){
        int N,M,x,y,res=0;
        Scanner in=new Scanner(System.in);
        N=in.nextInt();
        M=in.nextInt();
        int[]dp=new int[N+1];//dp[i]:以i为终点的最长路径
        int[] visit=new int[N+1];
        HashMap<Integer,ArrayList<Integer>>reversed_edges=new HashMap<>();
        visit[0]=1;
        for(int i=1;i<=M;i++){
            x=in.nextInt();
            y=in.nextInt();
            if(!reversed_edges.containsKey(y))reversed_edges.put(y,new ArrayList<>());
            reversed_edges.get(y).add(x);
        }
        for(int i=1;i<=N;i++){
            if(reversed_edges.containsKey(i)){
            for(Integer m:reversed_edges.get(i)){
                dp[i]=Math.max(dp[i],cal(m,visit,dp,reversed_edges)+1);
            }
            }
        }
        for(int i=1;i<=N;i++)res=Math.max(res,dp[i]);
        System.out.println(res);
    }
}

