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 lev[2000000];
int vis[2000000];
set<int> s;
vector<vector<int>> g;
void bfs(int node) {
  queue<int> q;
  q.push(node);
  lev[node] = 1;
  while (!q.empty()) {
    int p = q.front();
    q.pop();
    if (vis[p]) continue;
    vis[p] = 1;
    for (auto j : g[p]) {
      if (!vis[j]) {
        q.push(j);
        lev[j] = lev[p] + 1;
        s.insert(lev[j]);
      }
    }
  }
}
int main() {
  long long int n, i, j = 0, m = 0, cou = 0, sum = 0, f = 0;
  long long int mx = INT_MIN, mn = INT_MAX, lz = 0, rz = 0, lo = 0,
                ro = INT_MAX, mi = 0;
  long long int mid, k, d, l, r;
  long long int x, y, t;
  cin >> n;
  s.insert(1);
  g.resize(n + 1);
  vector<int> a;
  for (i = 1; i <= n; i++) {
    int v;
    cin >> v;
    if (v != -1) {
      g[v].push_back(i);
      g[i].push_back(v);
    } else
      a.push_back(i);
  }
  for (i = 0; i < a.size(); i++) {
    bfs(a[i]);
  }
  cout << *--s.end();
}
