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.BufferedInputStream;
import java.util.*;
import static java.lang.Math.*;

public class C116C {

	private int dfs(int[] arr, int i) {
		int res = 0;
		while(arr[i] != -2) {
			i = arr[i];
			res++;
		}
		return res;
	}
	
	public void solve() throws Exception {
		int n = nextInt();
		int[] arr = new int[n];
		for (int i = 0; i < arr.length; i++) {
			arr[i] = nextInt() - 1;
		}
		int res = 0;
		for (int i = 0; i < arr.length; i++) {
			res = max(dfs(arr, i) + 1, res);
		}
		println(res);
	}

	// ------------------------------------------------------

	void debug(Object... os) {
		System.err.println(Arrays.deepToString(os));
	}

	void print(Object... os) {
		if (os != null && os.length > 0)
			System.out.print(os[0].toString());
		for (int i = 1; i < os.length; ++i)
			System.out.print(" " + os[i].toString());
	}

	void println(Object... os) {
		print(os);
		System.out.println();
	}

	BufferedInputStream bis = new BufferedInputStream(System.in);

	String nextWord() throws Exception {
		char c = (char) bis.read();
		while (c <= ' ')
			c = (char) bis.read();
		StringBuilder sb = new StringBuilder();
		while (c > ' ') {
			sb.append(c);
			c = (char) bis.read();
		}
		return new String(sb);
	}

	String nextLine() throws Exception {
		char c = (char) bis.read();
		while (c <= ' ')
			c = (char) bis.read();
		StringBuilder sb = new StringBuilder();
		while (c != '\n' && c != '\r') {
			sb.append(c);
			c = (char) bis.read();
		}
		return new String(sb);
	}

	int nextInt() throws Exception {
		return Integer.parseInt(nextWord());
	}

	long nextLong() throws Exception {
		return Long.parseLong(nextWord());
	}

	public static void main(String[] args) throws Exception {
		new C116C().solve();
	}
}
