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 
#!/usr/bin/env python2.6
# coding=utf-8
import sys
import re

def compute(n, p):
    sub = [ set() for i in range(n) ]
    d = [ 0 for i in range(n) ]
    que = []
    for e, m in enumerate(p):
        if m == -2:
            que.append(e)
            d[e] = 1
        else:
            sub[m].add(e)
    while que:
        v = que[0]
        que = que[1:]
        for e in sub[v]:
            d[e] = d[v]+1
            que.append(e)
    return max(d)

if __name__ == "__main__":
    n = int(sys.stdin.readline().strip())
    p = map(lambda x:int(x)-1, [ sys.stdin.readline().strip() for i in range(n)])
    print("{0}".format(compute(n, p)))
        
    
