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?
The first line contains integer *n* (1 ≤ *n* ≤ 2000) — the number of employees.

The next *n* lines contain the integers *p*<sub class="lower-index">*i*</sub> (1 ≤ *p*<sub class="lower-index">*i*</sub> ≤ *n* or *p*<sub class="lower-index">*i*</sub> = -1). Every *p*<sub class="lower-index">*i*</sub> denotes the immediate manager for the *i*-th employee. If *p*<sub class="lower-index">*i*</sub> 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 (*p*<sub class="lower-index">*i*</sub> ≠ *i*). Also, there will be no managerial cycles.
5
-1
1
2
1
-1
3
4
-1
1
2
3
4
12
-1
1
2
3
-1
5
6
7
-1
9
10
11
4
6
-1
-1
2
3
1
1
3
3
-1
1
1
2
1
-1
1
2
2
-1
2
2
-1
-1
1
3
2
-1
1
3
3
-1
-1
-1
1
5
4
5
1
-1
4
3
12
-1
1
1
1
1
1
3
4
3
3
4
7
4
12
-1
-1
1
-1
1
1
5
11
8
6
6
4
5
12
-1
-1
-1
-1
-1
-1
-1
-1
2
-1
-1
-1
2
12
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
-1
1
12
3
4
2
8
7
1
10
12
5
-1
9
11
12
12
5
6
7
1
-1
9
12
4
8
-1
3
2
11
12
-1
9
11
6
6
-1
6
3
8
6
1
6
6
12
7
8
4
12
7
9
-1
-1
-1
8
6
-1
3
12
-1
10
-1
1
-1
5
9
12
-1
-1
3
-1
2
12
-1
7
9
12
1
7
-1
-1
8
5
4
-1
3
12
11
11
8
9
1
1
2
-1
10
3
-1
8
5
12
-1
8
9
-1
4
2
11
1
-1
6
-1
10
6
12
7
4
4
-1
6
7
9
7
-1
7
4
1
4
12
-1
6
10
6
1
11
5
9
11
5
1
9
4
12
8
6
6
8
6
-1
6
6
8
6
8
8
3
12
3
8
9
12
2
8
11
12
2
9
-1
11
7
7
7
6
-1
-1
4
4
-1
3
77
53
-1
48
-1
51
-1
54
43
70
19
64
31
45
58
6
61
41
56
34
-1
54
-1
66
65
74
16
38
1
2
61
-1
34
-1
-1
7
19
57
13
61
42
-1
34
65
16
31
61
6
5
34
15
-1
13
4
22
74
7
-1
70
38
61
45
53
46
22
-1
65
34
19
43
48
33
58
36
-1
47
24
34
6
5
2
3
4
5
-1
5
import sys
import math
import bisect
import heapq
import itertools
from itertools import accumulate
from sys import stdin,stdout
from math import gcd,floor,sqrt,log, ceil
from collections import defaultdict, Counter, deque
from bisect import bisect_left,bisect_right, insort_left, insort_right
    
mod=1000000007
    
def get_ints(): return map(int, sys.stdin.readline().strip().split())
def get_list(): return list(map(int, sys.stdin.readline().strip().split()))
def get_string(): return sys.stdin.readline().strip()
def get_int(): return int(sys.stdin.readline().strip())
def get_list_strings(): return list(map(str, sys.stdin.readline().strip().split()))
    
    
    
def solve():
    n = get_int()
    count = Counter()
    for i in range(n):
        num = get_int()
        if num in count:
            count[num].append(i+1)
        else:
            count[num] = [i+1]
    keys = sorted(count)
    ans = 0

    def dfs(count, start, val):
        nonlocal ans
        arr = []
        if start in count:
            arr = count[start]
            val += 1
        ans = max(val, ans)
        for num in arr:
            dfs(count, num, val)
    dfs(count, -1, 0)
    return ans
    
        


    

    
    
if __name__ == "__main__":
    
    print(solve())