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.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.Stack;
import java.util.StringTokenizer;
import static java.lang.Math.*;

public class A {

	public static void main(String[] args) {
		FastScanner sc =new FastScanner();
		PrintWriter pw = new PrintWriter(System.out);
		int n = sc.nextInt();
		Node[] nodes = new Node[n+1];
		for(int i = 0; i<= n; i++){
			nodes[i] = new Node();
		}
		ArrayList<Integer> root = new ArrayList<>();
		for(int i = 1; i <= n; i++){
			int p = sc.nextInt();
			if(p != -1){
				nodes[p].children.add(nodes[i]);
			}else{
				root.add(i);
			}
		}
		int ans = -1;
		for(int p : root){
			nodes[p].setDepth(0);
		}
		for(int i = 1; i <= n; i++){
			ans = max(ans, nodes[i].d);
		}
		pw.println(ans);	
		pw.close();
	}
	
	static class Node{
		int h, d;
		ArrayList<Node> children = new ArrayList<>();
		
		private int getHeight(){
			this.h = 0;
			for(Node child : children){
				this.h = max(this.h, child.getHeight());
			}
			return this.h+1;
		}

		private void setDepth(int parentDepth){
			this.d = parentDepth + 1;
			for(Node child : children){
				child.setDepth(this.d);
			}
		}
	}
	
	static void sort(int[] a) {
		ArrayList<Integer> l=new ArrayList<>();
		for (int i:a) l.add(i);
		Collections.sort(l);
		for (int i=0; i<a.length; i++) a[i]=l.get(i);
	}
	
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}

	
}

