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.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.regex.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;

public class Solution {

	// Complete the sockMerchant function below.
	
	static int [] array ;
	static boolean [] visited ;
	private static Scanner scanner = new Scanner(System.in);
	static int dfs(int i) {
		if(visited[i])return 0;
		visited[i] = true;
		
		if(array[i] == -1)return 1;
		
		return dfs(array[i] - 1) + 1;
		
	}
	public static void main(String[] args) throws IOException {
		class InputReader {
			BufferedReader reader;
			StringTokenizer tokenizer;

			public InputReader(InputStream stream) {
				reader = new BufferedReader(new InputStreamReader(stream));
				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 long nextLong() {
				return Long.parseLong(next());
			}

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

		}
		InputReader y = new InputReader(System.in);
		int n = y.nextInt();
		 array = new int [n];
		 visited = new boolean [n];
		 for (int i = 0; i < n; i++) {
			array [i] = y.nextInt();
		}
		 int ans = 0;
		 for (int i = 0; i < n; i++) {
			 for (int j = 0; j < visited.length; j++) {
				visited[j] = false;
			}
			 if(!visited[i]) {
				int x = dfs(i);
				if(x > ans) {
					ans = x ;
				}
			}
		}
		System.out.println(ans);
	}
}
