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.util.ArrayList;
import java.util.List;
import java.util.Scanner;
import java.util.Stack;

public class Party {
	
	static List<List<Integer>> adj;

	public static void main(String[] args) {
		
		Scanner in = new Scanner(System.in);
		
		
		int n = in.nextInt();
		
		adj = new ArrayList<>();
		
		for(int i=0; i<=n; i++) {
			adj.add(new ArrayList<>());
		}
		
		
		
		// NOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO
		//WRONG if forest > height of tree then forest count is max number of parties
		
		// we can put all forest people in same party
		
		int forest = 0;
		
		for(int i=1; i<=n; i++) {
			
			int parent = in.nextInt();
			if(parent == -1) {
				forest++;
				continue;
			}
			
			adj.get(parent).add(i);
		}
		
		
		Stack<Integer> stack = new Stack<>();
		
		boolean[] vis = new boolean[n+1];
		
		for(int i=1; i<adj.size(); i++) {
			if(!vis[i]) {
				topSort(i, vis, stack);
			}
		}
		
		vis = new boolean[n+1];
		int max = -1;
		
		while(!stack.isEmpty()) {
			int v = stack.pop();
			
			if(vis[v])
				continue;
			
			max = Math.max(max, height(v, vis));
		}
		
		System.out.println(max);
	}

	

	private static int height(int i, boolean[] vis) {
		vis[i] = true;
		
		
		if(adj.get(i).size() == 0)
			return 1;
		
		int maxHeight = 0;
		for(int k=0; k<adj.get(i).size(); k++) {
			maxHeight = Math.max(maxHeight, 1 +    height(adj.get(i).get(k), vis)        );
		}
		
		return maxHeight;
	}



	private static void topSort(int i, boolean[] vis, Stack<Integer> stack) {
		
		vis[i] = true;
		
		for(int k=0; k<adj.get(i).size(); k++) {
			int n = adj.get(i).get(k);
			if(!vis[n]) {
				topSort(n, vis, stack);
			}
		}
		
		stack.push(i);
		
	}

}
