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.Scanner;
import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int N = scan.nextInt();
        int M = scan.nextInt();
        //Build graph
        Map<Integer, List<Integer>> graph = new HashMap<>();
        for(int i = 0 ; i < M ; i++){
            int source = scan.nextInt();
            int target = scan.nextInt();
            graph.putIfAbsent(source, new ArrayList<>());
            graph.get(source).add(target);
        }

        int longestPath = 0;
        int[] longest = new int[N+1];
        Arrays.fill(longest, -1);
        for(int vertex = 1 ; vertex <= N ;vertex++){
            if(longest[vertex] == -1) {
                longestPath = Math.max(longestPath, dfs(graph, vertex, longest));
            }
        }
        System.out.println(longestPath);
    }

    private static int dfs(Map<Integer, List<Integer>> graph, Integer vertex, int[] cache) {
        if(cache[vertex] != -1) {
            return cache[vertex];
        }

        int longestFromHere = 0;
        if(graph.get(vertex) != null) {
            for(Integer neighbour : graph.get(vertex)) {
                longestFromHere = Math.max(longestFromHere, 1+dfs(graph, neighbour, cache));
            }
        }
        cache[vertex] = longestFromHere;
        return cache[vertex];
    }
}
