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.Scanner;

public class A115 {
	static class guy {
		guy boss;
		int dist = -1;

		public guy() {
			boss = null;
		}

		public int getDist() {
			if (dist == -1) {
				if (boss == null) {
					dist = 1;
				} else
					dist = 1 + boss.getDist();
			}
			return dist;
		}
	}

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		guy[] g = new guy[n];
		for (int x = 0; x < n; x++) {
			g[x] = new guy();
		}
		for(int x = 0;x<n;x++) {
			int next = in.nextInt();
			if(next==-1) continue;
			g[x].boss = g[next-1];
		}
		int max = 0;
		for(int x =0;x<n;x++ ){
			int k = g[x].getDist();
			if(k>max) max = k;
		}
		System.out.println(max);
	}
}
