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 
#include <bits/stdc++.h>
using namespace std;
vector<int> v[2001];
int visited[2001];
int a[2001];
int j = 0;
void dfs(int i, int& maxcount, int count) {
  visited[i] = 1;
  count++;
  for (int k = 0; k < v[i].size(); k++) {
    if (v[i][k] != -1 && !visited[v[i][k]]) {
      if (count > maxcount) {
        maxcount = count;
        a[j] = maxcount;
      }
      dfs(v[i][k], maxcount, count);
    }
  }
}
int main() {
  int n, i, p;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> p;
    if (p != -1) {
      v[i].push_back(p - 1);
      v[p - 1].push_back(i);
    } else if (p == -1) {
      v[i].push_back(-1);
    }
  }
  int maxcount;
  for (i = 0; i < n; i++) {
    maxcount = 0;
    if (find(v[i].begin(), v[i].end(), -1) != v[i].end()) {
      dfs(i, maxcount, 0);
      j++;
    }
  }
  cout << *max_element(a, a + n) + 1;
}
