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.StringTokenizer;
import java.util.*;

public class Solution {
    private static final long mod = 1000000007;

    public static void main(String args[]) throws Exception {

        FastScanner sc = new FastScanner();
        PrintWriter out=new PrintWriter(System.out);
        int t;
        t = 1;//sc.nextInt();

        while (t-- > 0) {
            solve(sc,out);
        }
        out.close();

    }

    public static void solve(FastScanner sc, PrintWriter out) {
        int n = sc.nextInt();
        ArrayList<Integer>[] al = new ArrayList[n+1];
        for(int i = 1;i<=n;i++) al[i] = new ArrayList<>();

        int[] ar = new int[n+1];

        for(int i = 1;i<=n;i++){
            int sup = sc.nextInt();
            if(sup != -1)
                al[sup].add(i);

        }

        for(int i = 1;i<=n;i++)
            dfs(i,ar,al);

        out.println(Arrays.stream(ar).max().orElse(0));
    }

    public static int dfs(int r, int [] ar, ArrayList<Integer> al[]){
        if(ar[r] != 0) return ar[r];

        for(int i : al[r])
            ar[r] = Math.max(dfs(i,ar,al), ar[r]);

        ar[r] = ar[r] + 1;
        return ar[r];
    }

    static class FastScanner {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");

        public String next() {
            while (!st.hasMoreElements())
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

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

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

        public int[] readArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = nextInt();
            }
            return a;
        }

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

    }
}