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.Arrays;
import java.util.Scanner;

public class C87 {
	public static void main(String[] args) {
//		System.out.println(2000);
//		System.out.println(-1);
//		for(int i = 1; i < 2000; i++) {
//			System.out.println(i);
//		}

		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int[] s = new int[n];
		int[] c = new int[n];
		int[][] r = new int[n][n];
		ArrayList<Integer> roots = new ArrayList<Integer>();
		for (int i = 0; i < n; i++) {
			s[i] = in.nextInt() - 1;
			if (s[i] != -2) {
				r[s[i]][c[s[i]]] = i;
				c[s[i]]++;
			}    else roots.add(i);
		}
//		long t1 = System.currentTimeMillis();
		int gc = 0;
		boolean end = false;
		int k = 0;
		while (roots.size()!=0) {
			ArrayList<Integer> nroots = new ArrayList<Integer>();
//			for(int i = 0; i < n; i++) {
//			    if (s[i]==-3) s[i] = -2;
//			}
			for (int i :roots) {
					for (int j = 0; j < c[i]; j++) {
						nroots.add(r[i][j]);
					}
			}
			roots = nroots;
			gc++;
		}
		System.out.println(gc);
//		System.out.println(System.currentTimeMillis()-t1);
	}


}
