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;
int n;
vector<int> tree[2010];
int dfs(int pos) {
  int ret = 0;
  for (int i = 0; i < tree[pos].size(); i++)
    ret = max(ret, 1 + dfs(tree[pos][i]));
  return ret;
}
int main() {
  scanf("%d", &n);
  vector<int> allah;
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    if (x == -1)
      allah.push_back(i);
    else {
      x--;
      tree[x].push_back(i);
    }
  }
  int ans = 0;
  for (int i = 0; i < allah.size(); i++) ans = max(ans, 1 + dfs(allah[i]));
  printf("%d\n", ans);
}
