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.Arrays;
import java.util.List;
import java.util.Scanner;

public class Main {

    List<Integer>[] list;
    int[] dp;
    void run() {
        Scanner sc = new Scanner(System.in);

        int n = sc.nextInt();
        int m = sc.nextInt();
        list = new List[n];
        for (int i = 0; i < n; i++) {
            list[i] = new ArrayList<>();
        }

        for (int i =0 ; i< m; i++) {
            int x = sc.nextInt() - 1;
            int y = sc.nextInt() - 1;
            list[x].add(y);
        }
        dp = new int[n];
        Arrays.fill(dp, -1);
        // dp[i] ... ノードiからスタートする場合の最大経路長
        int ans = -1;
        for (int i = 0; i < n; i++) {
            ans = Math.max(ans, rec(i));
        }
        System.out.println(ans);
    }

    int rec(int i) {
        if (dp[i] != -1) {
            return dp[i];
        }
        int res = 0;
        for (int j : list[i]) {
            res = Math.max(res,rec(j) + 1);
        }
        
        dp[i] = res;
        return res;
    }


    void debug(Object... os) {
        System.err.println(Arrays.deepToString(os));
    }

    public static void main(String[] args) {
        new Main().run();
    }
}
