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.ArrayList;
import java.util.Scanner;

public class Main {
    static ArrayList<ArrayList<Integer>> graph;
    static int[] indegree;
    static int[] dist;
    static boolean[] visited;

    public static void main(String[] args) {
        int n, m;
        Scanner sc = new Scanner(System.in);
        n = sc.nextInt();
        m = sc.nextInt();
        graph = new ArrayList<ArrayList<Integer>>(n);
        indegree = new int[n];
        dist = new int[n];
        visited = new boolean[n];
        for (int i = 0; i < n; i++)
            graph.add(i, new ArrayList<>());
        for (int i = 0; i < m; i++) {
            int a = sc.nextInt();
            int b = sc.nextInt();
            graph.get(a-1).add(b-1);
            indegree[b-1]++;
        }
        for (int i = 0; i < n; i++) {
            if (!visited[i] && indegree[i] == 0) {
                dfs(i);
            }
        }
        int ans = 0;
        for(int dis : dist) {
            ans = Math.max(ans, dis);
        }
        System.out.println(ans);

    }

    private static void dfs(int a) {
        if(visited[a]) return;
        visited[a] = true;
        for (int b : graph.get(a)) {
            dist[b] = Math.max(dist[b],dist[a] + 1);
            indegree[b]--;
            if (indegree[b] == 0) {
                dfs(b);
            }
        }
    }
}
