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.*;
import java.math.*;
import java.lang.*;
public class Main {
    static void dfs(int i, ArrayList<Integer>[] g, boolean[] vis, int[] dp) {
        vis[i]=true;
        for(int j:g[i]) {
            if(!vis[j])
                dfs(j,g,vis,dp);
            dp[i]=Math.max(dp[i],dp[j]+1);
        }
    }
    public static void main(String[] args) throws Exception {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw=new PrintWriter(System.out);
        String s=br.readLine();
        String[] c=s.split(" ");
        int n=Integer.valueOf(c[0]),
            m=Integer.valueOf(c[1]);
        ArrayList<Integer>[] g=new ArrayList[n];
        for(int i=0;i<n;i++)
            g[i]=new ArrayList<>();
        for(int i=0;i<m;i++) {
            s=br.readLine();
            c=s.split(" ");
            int x=Integer.valueOf(c[0])-1,
                y=Integer.valueOf(c[1])-1;
            g[x].add(y);
        }
        int[] dp=new int[n];
        boolean[] vis=new boolean[n];
        for(int i=0;i<n;i++) {
            if(!vis[i])
                dfs(i,g,vis,dp);
        }
        int ans=0;
        for(int i=0;i<n;i++)
            ans=Math.max(ans,dp[i]);
        pw.print(ans);
        br.close();
        pw.close();
    }
}