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 
#def dfs(a,i,h):
#    if a[i] != []:
#        height = []
#        for item in a[i]:
#            height.append(dfs(a,item,h+1))
#        return max(height)
#    else:
#        return h
#if __name__ == '__main__':
#    n = int(input())
#    root = []
#    ans = 0
#    a = [[] for i in range(n+1)]
#    for i in range(1,n+1):
#        temp = int(input())
#        if temp == -1:
#            root.append(i)
#        else:
#            a[temp].append(i)
#    h = []
#    for item in root:
#        h.append(dfs(a,item,1))
#    print(max(h))
n = int(input())
a = [0]
for i in range(1,n+1):
    a.append(int(input()))
ans = 0
for i in range(1,n+1):
    tmp = 0
    item = a[i]
    while item != -1:
        tmp += 1
        item = a[item]
    ans = max(ans,tmp)
print(ans+1)