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.*;


public class C{     

        private BufferedReader in;  
        private StringTokenizer st;
        
        void solve() throws IOException{
            
            int n = nextInt();
            int []tree = new int[n];
            int i = 0;
            while(n -- >0){
                tree[i++] = nextInt()-1;                
            }
            int max = 0;
            for (int j = 0; j < tree.length; j++) {
                int k = tree[j];
                int lev = 1;
                while(k != -2){
                    k = tree[k];
                    lev++;
                }
                max = Math.max(max, lev);
            }
            System.out.println(max);
        }
            

        C() throws IOException {
            in = new BufferedReader(new InputStreamReader(System.in));          
            eat("");
            solve();            
        }

        private void eat(String str) {
            st = new StringTokenizer(str);
        }

        String next() throws IOException {
            while (!st.hasMoreTokens()) {
                String line = in.readLine();
                if (line == null) {
                    return null;
                }
                eat(line);
            }
            return st.nextToken();
        }

        int nextInt() throws IOException {
            return Integer.parseInt(next());
        }

        long nextLong() throws IOException {
            return Long.parseLong(next());
        }

        double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }

        public static void main(String[] args) throws IOException {
            new C();
        }

        int gcd(int a,int b){
            if(b>a) return gcd(b,a);
            if(b==0) return a;
            return gcd(b,a%b);
        }

}
