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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.Scanner;
import java.math.*;
import java.util.StringTokenizer;

/**
 *
 * @author T4reQ
 */
public class Codeforces {

    static int[] parent;
    static boolean[] seen;
    static int depth = -1;

    public static void main(String[] args) throws Exception {
        FastScanner in = new FastScanner();
        int n = in.nextInt();
        parent = new int[n];
        seen = new boolean[n];

        for (int i = 0; i < n; i++) {
            int x = in.nextInt();
            if (x != -1) {
                parent[i] = x - 1;
            } else {
                parent[i] = -1;
            }
        }

        for (int i = 0; i < n; i++) {
            Arrays.fill(seen, false);
            dfs(i, 1);
        }
        System.out.println(depth);
    }

    static void dfs(int node, int len) {
        if (seen[node] || parent[node] == -1) {
            depth = Math.max(depth, len);
            return;
        }
        seen[node] = true;
        dfs(parent[node], len + 1);
    }

    static class FastScanner {

        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public FastScanner() {
            reader = new BufferedReader(new InputStreamReader(System.in), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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

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

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

        public String nextLine() {
            try {
                return reader.readLine();
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }

    }
}
