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<long long> a[(long long)2e5 + 1];
bool ch[(long long)2e5 + 1] = {false};
long long s;
bool f = true;
long long level[(long long)2e5 + 1];
void bfs(long long i) {
  queue<long long> q;
  q.push(i);
  level[i] = 1;
  ch[i] = true;
  while (!q.empty()) {
    long long o = q.front();
    q.pop();
    for (long long j = 0; j < a[o].size(); j++) {
      if (ch[a[o][j]] == false) {
        ch[a[o][j]] = true;
        level[a[o][j]] = level[o] + 1;
        q.push(a[o][j]);
      }
    }
  }
}
int main() {
  long long i, j, n, k, l, m;
  cin >> n;
  vector<long long> d;
  for (i = 1; i <= n; i++) {
    cin >> l;
    if (l == -1) {
      d.push_back(i);
      continue;
    }
    a[l].push_back(i);
  }
  long long ma = 0;
  for (i = 0; i < d.size(); i++) {
    for (j = 1; j <= n; j++) {
      ch[j] = false;
    }
    bfs(d[i]);
    for (j = 1; j <= n; j++) {
      ma = max(ma, level[j]);
    }
  }
  cout << ma;
}
