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 sc = new Scanner(System.in);

        Map<Integer, List<Integer>> graph = new HashMap<>();
        int n = sc.nextInt();
        int m = sc.nextInt();
        for (int i = 0; i < m; i++) {
            int x = sc.nextInt();
            int y = sc.nextInt();
            List<Integer> edges = graph.computeIfAbsent(x, k -> new ArrayList<>());
            edges.add(y);
        }

        Integer[] maxes = new Integer[n+1];
        int maxPath = 0;
        for (int i = 1; i <= n; i++) {
            maxPath = Math.max(maxPath, maxPath(graph, i, maxes));
        }

        System.out.println(maxPath);
    }

    private static int maxPath(Map<Integer, List<Integer>> graph, int i, Integer[] maxes) {
        List<Integer> edges = graph.get(i);
        if(edges == null || edges.isEmpty()) {
            return 0;
        }

        if(maxes[i] != null) {
            return maxes[i];
        }

        int max = 0;
        for (int edge: edges) {
            int path = maxPath(graph, edge, maxes);
            max = Math.max(path, max);
        }

        max++;
        maxes[i] = max;
        return max;
    }
}