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.*;

public class Main {

    public static void main(String args[]) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int m = in.nextInt();
        List<List<Integer>> graph = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            graph.add(new ArrayList<>());
        }
        for (int i = 0; i < m; i++) {
            int temp1 = in.nextInt();
            int temp2 = in.nextInt();
            graph.get(temp1 - 1).add(temp2-1);
        }
        int max = 0;
        //  System.out.println(go(mat,0,0));
        boolean vis[] = new boolean[n];
        int dis[] = new int[n];
        for (int i = 0; i < n; i++) {
            if (!vis[i]) {
                int temp = go(graph, i, 0, vis,dis);
                //System.out.println(i+" "+temp+" ");
                max = Math.max(max, temp);
            }
        }
        System.out.println(max);

    }

    static int go(List<List<Integer>> graph, int curr, int level, boolean vis[],int dis[]) {
        // System.out.println("Curr = "+curr+" "+level);
        int currL = level;
        if(dis[curr]>0){
            vis[curr] = true;
            return dis[curr];
        }
        for (int i = 0; i < graph.get(curr).size(); i++) {
            level = Math.max(level, go(graph, graph.get(curr).get(i), currL, vis,dis) + 1);
        }
        dis[curr] = level;
        vis[curr] = true;
        return level;
    }

}