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 Main {

    void walkMax(int[] p, int[] v, int i) {
        int d = 1;
        
        while(i >= 0) {
            v[i] = Math.max(v[i], d++);
            i = p[i];
        }
    }

    void solve() throws IOException {
        int n = nextInt();
        int[] p = new int[n];
        int[] v = new int[n];

        for(int i = 0; i < n; ++i)
            p[i] = nextInt() - 1;

        for(int i = 0; i < n; ++i)
            walkMax(p, v, i);

        int ans = 0;
        for(int i = 0; i < n; ++i)
            if(v[i] > ans)
                ans = v[i];

        out.println(ans);
    }

    void run() throws IOException {
        in = new BufferedReader(new InputStreamReader(System.in));
        out = System.out;
        solve();
        in.close();
        //out.close();
    }

    BufferedReader in;
    PrintStream out;
    StringTokenizer tok;

    void take() throws IOException {
        tok = new StringTokenizer(in.readLine());
    }

    String nextToken() throws IOException {
        while(tok == null || !tok.hasMoreTokens())
            take();

        return tok.nextToken();
    }

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

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

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

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