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.*;
import java.util.*;

public class Main {

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

	private long solve() {

		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int M = sc.nextInt();
		Map<Integer, List<Integer>> graph =new HashMap<>();
		for(int i=1;i<=M;i++){
			int u = sc.nextInt();
			int v = sc.nextInt();
			List<Integer> children = graph.getOrDefault(u, new ArrayList<Integer>());
			children.add(v);
			graph.put(u, children);
		}
		long [] dp = new long[N+1];
		long ans =0;
		for(int i=1;i<=N;i++){
			ans = Math.max(ans, solve(graph, i, dp));
		}

		return ans;
	}

	private long solve(Map<Integer, List<Integer>> graph, int u, long []dp) {
		if (!graph.containsKey(u))  return 0;
		if (dp[u]!=0) return dp[u];
		long bestChild =0;
		for(int child: graph.get(u)) {
			bestChild = Math.max(bestChild, solve(graph, child, dp));
		}
		dp[u] = bestChild+1;
		return dp[u];
	}



}