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 

class Tree:
    def __init__(self,rootObj):
        self.key = rootObj
        self.children = []
        
    def newChild(self,newNode):
        self.children.append(newNode)

    def getChildren(self):
        return self.children

    def setRootVal(self,obj):
        self.key = obj

    def getRootVal(self):
        return self.key
 



t = int(input())
nodes=[Tree(i) for i in range(t)]
noManager = []

for i in range(t):
    val = int(input())
    if val == -1:
        noManager.append(i)
    else:
        nodes[val-1].newChild(nodes[i])
        
def maxDepth(root): 
    if root is None: 
        return 0
    q = [] 
    q.append(root) 
    height = 0 
    while(True): 
        nodeCount = len(q) 
        if nodeCount == 0 : 
            return height  
        height += 1 
        while(nodeCount > 0): 
            node = q[0] 
            q.pop(0) 
            for child in node.getChildren():
                q.append(child)
            nodeCount -= 1
    
    
print(max([maxDepth(nodes[i]) for i in noManager]))
        
            
    