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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.StringTokenizer;

public class Main
{
	
	static int[] dp;
	static ArrayList<Integer>[] graph;
	public static int dfs(int u) {
		if(dp[u]!=-1) {
			return dp[u];
		}
		int ans = 0;
		for(int v: graph[u]) {
			ans = Math.max(ans, dfs(v)+1);
		}
		dp[u] = ans;
		return ans;
	}
	
	public static void main(String[] args) throws IOException {
		
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(br.readLine());
		
		int N = Integer.parseInt(st.nextToken()); int M = Integer.parseInt(st.nextToken());
		
		graph = new ArrayList[N+1];
		for(int i=1; i<=N; i++) {
			graph[i] = new ArrayList<Integer>();
		}
		
		for(int i=0; i<M; i++) {
			st = new StringTokenizer(br.readLine());
			graph[Integer.parseInt(st.nextToken())].add(Integer.parseInt(st.nextToken()));
		}
		
		dp = new int[N+1];
		Arrays.fill(dp, -1);
		
		int ans = 0;
		for(int i=1; i<=N; i++) {
			ans = Math.max(ans, dfs(i));
		}
		System.out.println(ans);
	}
}
