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.io.*;

class Main {

	private static void updateDistances(List<Integer> []adjList, int []dp, boolean []visited, int src) {
		
		visited[src] = true;

		for(int target: adjList[src]) {
			if (!visited[target])
				updateDistances(adjList, dp, visited, target);

			dp[src] = Math.max(dp[src], 1 + dp[target]);
		}
	}

	public static void main(String[] args) throws Exception{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        
        String []s = br.readLine().split(" ");

		int N = Integer.parseInt(s[0]);
		int M = Integer.parseInt(s[1]);

		List<Integer> []adjList = new ArrayList[N];

		for(int i=0; i < N; ++i)
			adjList[i] = new ArrayList<>();


		boolean []visited = new boolean[N];

		int []dp = new int[N];
		Arrays.fill(dp, 0);

		for(int i=0; i < M; ++i) {
			s = br.readLine().split(" ");

			int u = Integer.parseInt(s[0]) - 1;
			int v = Integer.parseInt(s[1]) - 1;

			adjList[u].add(v);
		}

		for(int i=0; i < N; ++i)
			if (!visited[i])
				updateDistances(adjList, dp, visited, i);

		int result = 0;
		for(int i=0; i < N; ++i)
			result = Math.max(result, dp[i]);

		System.out.println(result);
    }
}