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.*;
import java.util.*;

import static java.lang.Math.*;
import static java.util.Arrays.*;

public class ladders115a {
    static BufferedReader __in;
    static PrintWriter __out;
    static StringTokenizer input;

    public static void main(String[] args) throws IOException {
        __in = new BufferedReader(new InputStreamReader(System.in));
        __out = new PrintWriter(new OutputStreamWriter(System.out));

        int n = readInt();
        int ans = 0;
        Node[] g = new Node[n];
        for(int i = 0; i < n; ++i) {
            g[i] = new Node();
        }
        for(int i = 0; i < n; ++i) {
            int p = readInt();
            if(p > 0) {
                g[i].next = g[p - 1];
            }
        }
        for(int i = 0; i < n; ++i) {
            ans = max(ans, g[i].d());
        }
        println(ans);

        close();
    }

    static class Node {
        Node next;
        int d = -1;

        int d() {
            if(d > 0) {
                return d;
            }
            if(next == null) {
                return 1;
            }
            return d = (1 + next.d());
        }
    }

    static final int IMAX = 2147483647;
    static final long LMAX = 9223372036854775807L;
    static void read() throws IOException {input = new StringTokenizer(__in.readLine());}
    static String readLine() throws IOException {return __in.readLine();}
    static int nextInt() {return Integer.parseInt(input.nextToken());}
    static int readInt() throws IOException {return Integer.parseInt(__in.readLine());}
    static void print(String s) {__out.print(s);}
    static void println(String s) {__out.println(s);}
    static void print(Object o) {__out.print(o);}
    static void println(Object o) {__out.println(o);}
    static void printy() {__out.println("YES");}
    static void printn() {__out.println("NO");}
    static void printyn(boolean b) {__out.println(b ? "YES" : "NO");}
    static void println(int[] a) {for(int i = 0, len = a.length - 1; i < len; __out.print(a[i]), __out.print(' '), i++); __out.println(a[a.length - 1]);}
    static void println() {__out.println();}
    static void close() {__out.close();}
}