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;
const int INF = (int)2E9;
const long long LINF = (long long)9E18;
const long double PI = acos(-1.0);
const long double EPS = 1E-11;
long long n;
long long a[2001], color[2001], bac[2001], sl[2001];
long long ke[2001][2001];
void doc() {
  long long i, u;
  cin >> n;
  for (int i = 1; i <= (n); ++i) {
    cin >> u;
    if (u != -1) {
      sl[u]++;
      ke[u][sl[u]] = i;
    }
  }
}
void dfs(long long u) {
  long long i, v;
  color[u] = 1;
  for (i = 1; i <= sl[u]; i++) {
    v = ke[u][i];
    bac[v] = bac[u] + 1;
    dfs(v);
  }
}
void tinh() {
  long long i, kq;
  for (int i = 1; i <= (n); ++i) {
    if (color[i] == 0) {
      bac[i] = 1;
      dfs(i);
    }
  }
  kq = 0;
  for (int i = 1; i <= (n); ++i) {
    if (bac[i] > kq) kq = bac[i];
  }
  cout << kq;
}
int main() {
  doc();
  tinh();
  return 0;
}
