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.*;
import java.lang.*;

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

		final int max_node = 100005;

		int[] dist = new int[max_node]; 
		ArrayList<Integer>[] edges = new ArrayList[max_node];
		int[] in_degree = new int[max_node];
		boolean[] visited = new boolean[max_node];

		for (int i = 0; i < max_node; i++) { 
			edges[i] = new ArrayList<Integer>(); 
		}

	//	System.out.println("Enter "+m+" edges");
		for(int i = 0; i < m ; ++i){
			int x = in.nextInt();
			int y = in.nextInt();
			in_degree[y]++;

			edges[x].add(y);
		}

		Main graph = new Main();

		for(int i =1; i<=n; i++){
			if(visited[i] == false && in_degree[i] == 0){
				graph.dfs(i,edges, visited, in_degree, dist);
			}
		}

		int answer = 0;
		for(int i = 1; i < max_node; ++i){
			if(dist[i] > answer){
				answer = dist[i];
			}
		}
		
		System.out.println(answer);


	}

	private void dfs(int node,ArrayList<Integer>[] edges, boolean[] visited, int[] in_degree, int[] dist){
		if(visited[node] == true){
			return;
		}
		visited[node] = true;
		for(int edge : edges[node]){
			in_degree[edge]--;
			dist[edge] = Math.max(dist[edge], dist[node]+1);
			if(in_degree[edge] == 0 ){
				dfs(edge, edges, visited, in_degree, dist);
			}
		}
	}

}