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.math.BigInteger;
import java.util.Arrays;
import java.util.Comparator;
import java.util.LinkedList;


public class Prac2 {

static class UnionFind{
     int[] arr;
     int[] size;
     public UnionFind(int n){
        arr = new int[n + 1];
        size = new int[n + 1];

        for(int i = 0; i <= n ; ++i){
            arr[i] = -1;
            size[i] = 1;
        }
     }

    public boolean union(int k , int j){
        int pk = find(k);
        int pj = find(j);
        if(pk != pj){
            if(size[pk] > size[pj]){
                arr[pj] = pk;
                size[pk] += size[pj];
                size[pj] = 0;
            }else{
                arr[pk] = pj;
                size[pj]+= size[pk];
                size[pk] = 0;
            }
            return true;
        }else{
            return false;
        }
    }

    public int find(int k){
        if(arr[k] != -1){
            while(arr[arr[k]] != -1){
                arr[k] = arr[arr[k]];
            }
            return arr[k];
        }else{
            return k;
        }
    }

    public int[] getArray(){
        return arr;
    }

    public int[] getSizeArray(){
        return size;
    }
}
    static int length = 0;
    public static void main(String[] ar) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringBuilder sb = new StringBuilder();

        int n = pi(br.readLine());
        boolean[][] a = new boolean[n+1][n+1];
        int[] b = new int[n+1];
        for(int i = 1; i <= n ; ++i){
            int k = pi(br.readLine());
            if(k == -1)
            b[i] = k;
            else
            a[k][i] = true;
        }

//        int l = 0;
        for(int i = 1; i <=n; ++i){
            if(b[i] == -1){
                boolean[] visited = new boolean[n+1];
                dfs(i,a,visited, 1);
            }
        }
        System.out.println(length);
    }

    private static void dfs(int start, boolean[][] a, boolean[] visited,int l) {
        length = Math.max(l, length);
        visited[start] = true;
        for(int i = 1; i < a.length ; ++i){
            if(a[start][i] && !visited[i]){
                 dfs(i, a, visited,l+1);
            }
        }

    }

    public static String[] ti(BufferedReader br) throws IOException{
        return br.readLine().split(" ");
    }

    public static int pi(String s){
        return Integer.parseInt(s);
    }
}