A company has n employees numbered from 1 to n. Each employee either has no immediate manager or exactly one immediate manager, who is another employee with a different number. An employee A is said to be the superior of another employee B if at least one of the following is true:

  * Employee A is the immediate manager of employee B
  * Employee B has an immediate manager employee C such that employee A is the superior of employee C. 



The company will not have a managerial cycle. That is, there will not exist an employee who is the superior of his/her own immediate manager.

Today the company is going to arrange a party. This involves dividing all n employees into several groups: every employee must belong to exactly one group. Furthermore, within any single group, there must not be two employees A and B such that A is the superior of B.

What is the minimum number of groups that must be formed?

Input

The first line contains integer n (1 ≤ n ≤ 2000) — the number of employees.

The next n lines contain the integers pi (1 ≤ pi ≤ n or pi = -1). Every pi denotes the immediate manager for the i-th employee. If pi is -1, that means that the i-th employee does not have an immediate manager. 

It is guaranteed, that no employee will be the immediate manager of him/herself (pi ≠ i). Also, there will be no managerial cycles.

Output

Print a single integer denoting the minimum number of groups that will be formed in the party.

Examples

Input

5
-1
1
2
1
-1


Output

3

Note

For the first example, three groups are sufficient, for example: 

  * Employee 1 
  * Employees 2 and 4 
  * Employees 3 and 5 
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.StringTokenizer;

public class Main {

	final static long mod = 1000000007;

	public static void main(String[] args) throws Exception {

		STDIN scan = new STDIN();
		PrintWriter pw = new PrintWriter(System.out);
	
		int n = scan.nextInt();
		Node[] node = new Node[n];
		for(int i = 0; i < n; i++) node[i] = new Node();
		for(int i = 0; i < n; i++) {
			int par = scan.nextInt()-1;
			if(par < 0) node[i].depth = 1;
			else node[i].parent = node[par];
		}
		int max = Integer.MIN_VALUE;
		for(int i = 0; i < n; i++) max = Math.max(max, calcDepth(node[i]));
		pw.println(max);
		
		pw.flush();
	}
	
	static int calcDepth(Node node) {
		if(node.depth > 0) return node.depth;
		node.depth = 1 + calcDepth(node.parent);
		return node.depth;
	}
	
	static class Node{
		Node parent = null;
		int depth = -1;
	}

	static class STDIN {
		BufferedReader br;
		StringTokenizer st;

		public STDIN() {
			br = new BufferedReader(new InputStreamReader(System.in));
			st = null;
		}

		int nextInt() throws Exception {
			return Integer.parseInt(next());
		}

		long nextLong() throws Exception {
			return Long.parseLong(next());
		}

		double nextDouble() throws Exception {
			return Double.parseDouble(next());
		}

		String next() throws Exception {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		String nextLine() throws Exception {
			return br.readLine();
		}
	}
}