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.util.Scanner;
import java.util.Vector;

public class Main {
	static Vector<Integer> [] link;
	static int dp[];
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		link = new Vector[n+1];
		for (int i=1;i<=n;i++)
			link[i] = new Vector();
		dp = new int[n+1];
		
		for (int i=0;i<m;i++) {
			int a = sc.nextInt();
			int b = sc.nextInt();
			link[a].add(b);
		}
		int ans = 0;
		for (int i=1;i<=n;i++) {
			int l = f(i);
			if (ans<l) ans = l;
		}
		System.out.println(ans);
		
	}
	private static int f(int i) {
		if (dp[i]>0) return dp[i];
		int ans = 0;
		for (int j : link[i]) {
			int l = f(j) + 1;
			if (ans<l) ans = l;
		}
		dp[i] = ans;
		return ans;
	}
}