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.InputStream;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;

public class codeforces {
	
	public static void main(String[] args) {
		FS scan = new FS(System.in);
		int N = scan.nextInt();
		int[] h = new int[N+1];
		int[] par = new int[N+1];
		int max = 1;
		for(int i=1;i<=N;i++){
			int p = scan.nextInt();
			par[i] =p;
			if(p==-1)
				h[i]=1;
		}
		for(int i=1;i<=N;i++){
			for(int j=1;j<=N;j++){
				if(par[j]==-1)continue;
				if(h[par[j]]==0)continue;
				h[j]=h[par[j]]+1;
				max = Math.max(max, h[j]);
			}
		}
		System.out.println(max);
	}
	private static class Pair implements Comparable<Pair>{
		int a,b;
		public Pair(int a, int b) {
			this.a=a;
			this.b=b;
		}
		public int compareTo(Pair o) {
			if(this.a==o.a)return Integer.compare(b, o.b);
			else return Integer.compare(a,o.a);
		}
		
	}
	private static long exp(long a, long b, long mod) {
		if(b==0)return 1;
		long t = exp(a,b/2,mod);
		return (b%2==0) ? (t*t)%mod:(a*((t*t)%mod))%mod;
	}
	private static class FS {
		BufferedReader br;
		StringTokenizer st;
		public FS(InputStream in) {
			br = new BufferedReader(new InputStreamReader(in));
		}
		String next(){
			while(st==null||!st.hasMoreElements()){
				try{st = new StringTokenizer(br.readLine());}
				catch(IOException e){e.printStackTrace();}
			}
			return st.nextToken();
		}
		int nextInt() {return Integer.parseInt(next());}
		long nextLong() {return Long.parseLong(next());}
		double nextDouble() { return Double.parseDouble(next());}
	}
}
