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.*;

public class Main {
    static Stack<Integer> sort; static boolean[] visited; static ArrayList<Integer> [] adj;
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt(); int m = sc.nextInt();
        adj = new ArrayList[n+1];
        for (int i = 1; i <= n; i++) adj[i] = new ArrayList<>();
        for (int i = 0; i < m; i++) {
            int a = sc.nextInt(); int b = sc.nextInt();
            adj[a].add(b);
        }
        visited = new boolean[n+1]; ArrayList<Integer> order = new ArrayList<>(); sort = new Stack<>();
        for (int i = 1; i <= n; i++) {
            if (!visited[i]) {
                dfs(i);
            }
        }
        while (!sort.isEmpty()) order.add(sort.pop());
        int [] dp = new int[n+1]; ArrayList<Integer> [] rev = new ArrayList[n+1];
        for (int i = 1; i <= n; i++) rev[i] = new ArrayList<>();
        for (int i = 1; i <= n; i++) {
            for (Integer j: adj[i]) {
                rev[j].add(i);
            }
        }
        long ans = 0;
        for (int i = 0; i < order.size(); i++) {
            int here = order.get(i);
            for (Integer come: rev[here]) {
                dp[here] = Math.max(dp[here], 1 + dp[come]);
            }
            ans = Math.max(dp[here], ans);
        }
        System.out.println(ans);
    }

    static void dfs(int cur) {
        visited[cur] = true;
        for (Integer next: adj[cur]) {
            if (!visited[next]) {
                dfs(next);
            }
        }
        sort.push(cur);
    }

}