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 
function solve(input) {
    var companyMap = input.reduce(function (acc, curr, index) {
        if(curr !== -1) acc[curr] = (acc[curr] || []).concat(index)
        return acc
    }, {})

    var numbersOfLevels = Object.keys(companyMap)
        .filter(function(key) {
            return input[key] === -1
        })
        .map(function(key) {
            return count(companyMap, companyMap[key], 2)
        })
    return numbersOfLevels.length ? Math.max.apply(Math, numbersOfLevels) : 1
}

// function isTopLevelManager(input) {
//     return function () {
//
//     }
// }

function count(main, arr, c) {
    var res = arr
        .map(function(x) {
            if(!main[x]) return c
            var k = count(main, main[x], c + 1)
            return k
        })

    return Math.max.apply(Math, res)
}

var numOfEmployees = readline();
var nums = [-1]
while (numOfEmployees) {
    var next = Number(readline())
    nums.push(next)
    numOfEmployees--
}

print(solve(nums))