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 mod = 1000000007;
const int N = 3e5 + 5, M = N;
vector<int> g[2010];
bool vis[2020];
int level[2020];
int bfs(int src) {
  queue<int> q;
  q.push(src);
  vis[src] = 1;
  level[src] = 1;
  int mx = 1;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    for (int v : g[u]) {
      if (!vis[v]) {
        vis[v] = 1;
        level[v] = level[u] + 1;
        mx = max(mx, level[v]);
        q.push(v);
      }
    }
  }
  return mx;
}
void solve() {
  int n, m, u;
  scanf("%d", &n);
  vector<int> v;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &u);
    if (u != -1) {
      g[u].push_back(i);
      g[i].push_back(u);
    } else
      v.push_back(i);
  }
  int res = 0, sz = v.size();
  for (int i = 0; i < sz; i++) {
    res = max(res, bfs(v[i]));
  }
  printf("%d\n", res);
}
int main() {
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
