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 T;
int N, M;
string S;
vector<int> data;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  vector<int> adjList[2005];
  queue<int> q;
  bool visit[2005] = {false};
  int dist[2005];
  cin >> N;
  for (int i = int(1); i <= int(N); i++) {
    int v;
    cin >> v;
    if (v == -1) {
      q.push(i);
      visit[i] = true;
      dist[i] = 1;
    } else {
      adjList[v].push_back(i);
    }
  }
  int groupCnt = 1;
  while (!q.empty()) {
    int cur = q.front();
    q.pop();
    vector<int>& neighbors = adjList[cur];
    for (int i = int(0); i <= int(neighbors.size() - 1); i++) {
      int nxt = neighbors[i];
      if (visit[nxt]) continue;
      visit[nxt] = true;
      dist[nxt] = dist[cur] + 1;
      q.push(nxt);
      groupCnt = max(groupCnt, dist[nxt]);
    }
  }
  cout << groupCnt << endl;
  return 0;
}
