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.util.*;
public class Holidays {

	class Vertex {
		int num;
		boolean isVisited;
		Vertex(int a) {
			num = a;
			isVisited = false;
		}
	}
	Vertex[] vertexList;
	int[][] mat;
	Stack<Integer> stack = new Stack<Integer>();
	Scanner sc;
	int n;
	int c;
	int max = 1;
	ArrayList<Integer> boss = new ArrayList<Integer>();
	public void prepare() {
		sc = new Scanner(System.in);
		n = sc.nextInt();
		vertexList = new Vertex[n];
		mat = new int[n][n];
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < n; j++) {
				mat[i][j] = 0;
			}
		}
		for (int i = 0; i < n; i++) {
			vertexList[i] = new Vertex(i);
		}
	}
	public void addEdge(int st, int end) {
		mat[st][end] = 1;
		mat[end][st] = 1;
	}
	public void display(int v) {
		System.out.print(vertexList[v].num);
	}
	public void dfs(int k) {
		vertexList[k].isVisited = true;
		stack.push(k);
		
		while(!stack.isEmpty()) {
			int current = stack.peek();
			int v = getV(current);
			
			if (v == -1) {
				stack.pop();
				max = Math.max(max, c+1);
				c--;
			} else {
				vertexList[v].isVisited = true;
				stack.push(v);
				c++;
			}
		}
	}
	public int getV(int v) {
		for (int i = 0; i < n; i++) {
			if (mat[v][i] == 1 && vertexList[i].isVisited == false) 
				return i;
		}
		return -1;
	}
	public void go() {
		prepare();
		for (int i = 0; i < n; i++) {
			int a = sc.nextInt();
			if (a == -1) {
				boss.add(i);
				continue;
			}
			addEdge(i,a-1);
		}
		for(int a : boss) {
			dfs(a);
			c = 0;
		}
		System.out.println(max);
	}
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		new Holidays().go();
	}

}
