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 {
	Scanner sc = new Scanner(System.in);
	int n,m;
	List<List<Integer>> graph;
	static int[] memo;//graph.get(i)の有向パスの最大を記録

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

	private void solve() {
		n = nextInt();
		m = nextInt();
		graph = new ArrayList<List<Integer>>();
		for(int i=0;i<=n;i++)graph.add(new ArrayList<Integer>());
		for(int i=0;i<m;i++)graph.get(nextInt()).add(nextInt());

		memo = new int [n+1];
		Arrays.fill(memo, -1);
		for(int i=1;i<=n;i++)if(memo[i]<0)memo[i] = SearchVerticesPass(i);

		Arrays.sort(memo);
		System.out.println(memo[n]);
	}

	private int SearchVerticesPass(int vi) {
		if(memo[vi]>=0)return memo[vi];//
		List<Integer>V = graph.get(vi);
		int cntmax=0,t;
		for(int i=0;i<V.size();i++)if((t=SearchVerticesPass(V.get(i))+1)>cntmax)cntmax=t;
		memo[vi] = cntmax;
		return memo[vi];
	}

	private int nextInt() {
		return Integer.parseInt(sc.next());
	}
} 