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.io.*;
import java.math.*;
import java.text.*;
import java.util.*;
import java.util.regex.*;

public class Main {

    public static void dfs(ArrayList<Integer>[] graph, int n, int u, 
        boolean[] visited, int[] dp) {
        visited[u] = true;

        for(int v : graph[u]) {
            if(!visited[v]) {
                dfs(graph, n, v, visited, dp);
            }
            dp[u] = Math.max(dp[u], dp[v] + 1);
        }
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();

        ArrayList<Integer>[] graph = new ArrayList[n + 1];
        boolean[] visited = new boolean[n + 1];
        int[] dp = new int[n + 1];

        for(int i = 0; i <= n; i++) {
            graph[i] = new ArrayList<>();
        }

        for(int i = 0; i < m; i++) {
            int x = sc.nextInt();
            int y = sc.nextInt();
            graph[x].add(y);
        }

        for(int i = 1; i <= n; i++) {
            if(!visited[i]) {
                dfs(graph, n, i, visited, dp);
            }
        }

        int result = 0;
        for(int i = 1; i <= n; i++) {
            result = Math.max(result, dp[i]);
        }

        System.out.println(result);
  }
}
