Problem: 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

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.Scanner;

/**
 * Longest Path
 */
public class Main {

    public static void main(String[] args) throws Exception {
        try (Scanner sc = new Scanner(new BufferedReader(new InputStreamReader(System.in)))) {
            int N = sc.nextInt();
            int M = sc.nextInt();

            List<List<Integer>> g = new ArrayList<>();
            for (int i = 0; i < N; i++) {
                g.add(new ArrayList<>());
            }

            for (int i = 0; i < M; i++) {
                int x = sc.nextInt() - 1;
                int y = sc.nextInt() - 1;

                g.get(y).add(x);
            }

            int[] dp = new int[N];

            int max = 0;
            for (int i = 0; i < N; i++) {
                int d = solve(i, dp, g);

                max = Math.max(max, d);
            }

            System.out.println(max);
        }
    }

    static int solve(int x, int[] dp, List<List<Integer>> g) {
        if (dp[x] != 0) {
            return dp[x];
        }

        for (Integer e : g.get(x)) {
            dp[x] = Math.max(dp[x], solve(e, dp, g) + 1);
        }

        return dp[x];
    }

}