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[2005];
vector<int> head;
int maxd;
bool vis[2005];
void dfs(int x, int dep) {
  vis[x] = 1;
  if (dep > maxd) maxd = dep;
  for (vector<int>::iterator it = v[x].begin(); it != v[x].end(); it++) {
    if (vis[(*it)]) continue;
    dfs((*it), dep + 1);
  }
}
int main() {
  int n;
  while (~scanf("%d", &n)) {
    head.clear();
    for (int i = 1; i <= n; i++) {
      v[i].clear();
      vis[i] = 0;
    }
    for (int i = 1; i <= n; i++) {
      int x;
      scanf("%d", &x);
      if (x == -1) {
        head.push_back(i);
      } else {
        v[i].push_back(x);
        v[x].push_back(i);
      }
    }
    maxd = 0;
    for (vector<int>::iterator it = head.begin(); it != head.end(); it++) {
      dfs((*it), 1);
    }
    printf("%d\n", maxd);
  }
}
